Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 11 Geometric Progression

Question 1. Find, which of the following sequence form a G.P. :
(i) 8, 24, 72, 216, ……
(ii) \( \frac{1}{8}, \frac{1}{24}, \frac{1}{72}, \frac{1}{216} \), ……
(iii) 9, 12, 16, 24, ……
Answer:
Solution 1(i):
Given sequence: 8, 24, 72, 216……
Now, \( \frac{24}{8} = 3, \frac{72}{24} = 3, \frac{216}{72} = 3 \)
Since \( \frac{24}{8} = \frac{72}{24} = \frac{216}{72} = \ldots = 3 \), the given sequence is a G.P.
with common ratio 3.

Solution 1(ii):
Given sequence: \( \frac{1}{8}, \frac{1}{24}, \frac{1}{72}, \frac{1}{216} \), ……
Now, \( \frac{\frac{1}{24}}{\frac{1}{8}} = \frac{1}{3}, \frac{\frac{1}{72}}{\frac{1}{24}} = \frac{1}{3}, \frac{\frac{1}{216}}{\frac{1}{72}} = \frac{1}{3} \)
Since \( \frac{\frac{1}{24}}{\frac{1}{8}} = \frac{\frac{1}{72}}{\frac{1}{24}} = \frac{\frac{1}{216}}{\frac{1}{72}} = \ldots = \frac{1}{3} \), the given sequence is a G.P.
with common ratio \( \frac{1}{3} \).

Solution 1(iii):
Given sequence: 9, 12, 16, 24……
Now, \( \frac{12}{9} = \frac{4}{3}, \frac{16}{12} = \frac{4}{3}, \frac{24}{16} = \frac{3}{2} \)
Since \( \frac{24}{16} = \frac{72}{24} = \frac{216}{72} \), the given sequence is not a G.P.
In simple words: A G.P. has the same common ratio between every two consecutive terms. Only sequences (i) and (ii) have this property.

📝 Teacher's Note: Tell students to check if the ratio between each pair of consecutive terms is the same. If yes, it's a G.P. If not, it's not a G.P.

🎯 Exam Tip: Always calculate at least three ratios to check. Write "common ratio = ..." clearly. This gets you full marks.

Question 2. Find the 9th term of the series : 1, 4, 16, 64 ……
Answer:
Given sequence: 1, 4, 16, 64……
Now, \( \frac{4}{1} = 4, \frac{16}{4} = 4, \frac{64}{16} = 4 \)
Since \( \frac{4}{1} = \frac{16}{4} = \frac{64}{16} = \ldots = 4 \), the given sequence is a G.P.
with first term, a = 1 and common ratio, r = 4.
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow t_9 = 1 \times 4^8 = 65536 \)
In simple words: We multiply the first term by the common ratio raised to the power (n-1) to get the nth term.

📝 Teacher's Note: Show students the formula \( t_n = ar^{n-1} \). Make them practice substituting values carefully. Common mistake is using n instead of (n-1).

🎯 Exam Tip: Write the formula first. Then substitute a = 1, r = 4, n = 9. Calculate \( 4^8 \) step by step to avoid errors.

Question 3. Find the seventh term of the G.P. : 1, \( \sqrt{3} \), 3, \( 3\sqrt{3} \) ……
Answer:
Given G.P.: 1, \( \sqrt{3} \), 3, \( 3\sqrt{3} \), …
Here,
First term, a = 1
Common ratio, r = \( \frac{\sqrt{3}}{1} = \sqrt{3} \)
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow t_7 = 1 \times (\sqrt{3})^6 = 27 \)
In simple words: When we have square roots in a G.P., we find the common ratio by dividing consecutive terms. Then use the same formula.

📝 Teacher's Note: Help students see that \( (\sqrt{3})^6 = (\sqrt{3})^2 \times (\sqrt{3})^2 \times (\sqrt{3})^2 = 3 \times 3 \times 3 = 27 \).

🎯 Exam Tip: When you see square roots, be very careful with powers. Write \( (\sqrt{3})^6 = (3^{1/2})^6 = 3^3 = 27 \) to show your working.

Question 4. Find the 8th term of the sequence : \( \frac{3}{4}, 1\frac{1}{2}, 3 \), ……
Answer:
Given sequence: \( \frac{3}{4}, 1\frac{1}{2}, 3 \), ……
i.e. \( \frac{3}{4}, \frac{3}{2}, 3 \), ……
Now, \( \frac{\frac{3}{2}}{\frac{3}{4}} = 2, \frac{3}{\frac{3}{2}} = 2 \)
Since \( \frac{\frac{3}{2}}{\frac{3}{4}} = \frac{3}{\frac{3}{2}} = \ldots = 2 \), the given sequence is a G.P.
with first term, a = \( \frac{3}{4} \) and common ratio, r = 2.
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow t_8 = \frac{3}{4} \times 2^7 = \frac{3}{4} \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 3 \times 2^5 = 96 \)
In simple words: Convert mixed numbers to improper fractions first. Then find the common ratio and apply the formula.

📝 Teacher's Note: Always convert mixed numbers like \( 1\frac{1}{2} \) to improper fractions \( \frac{3}{2} \) before solving. This prevents errors.

🎯 Exam Tip: Show the conversion clearly: \( 1\frac{1}{2} = \frac{3}{2} \). Then calculate \( 2^7 = 128 \) step by step to get the final answer.

Question 5. Find the 10th term of the G.P. : 12, 4, \( 1\frac{1}{3} \), ……
Answer:
Given G.P.: 12, 4, \( 1\frac{1}{3} \), ……
Here,
First term, a = 12
Common ratio, r = \( \frac{4}{12} = \frac{1}{3} \)
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow t_{10} = 12 \times \left(\frac{1}{3}\right)^9 = 12 \times \frac{1}{19683} = \frac{4}{6561} \)
In simple words: When the common ratio is a fraction less than 1, the terms get smaller and smaller as we go further.

📝 Teacher's Note: Show students that \( \left(\frac{1}{3}\right)^9 = \frac{1}{3^9} \). Help them calculate \( 3^9 = 19683 \) step by step.

🎯 Exam Tip: When r is a fraction, be careful with powers. Write \( \left(\frac{1}{3}\right)^9 = \frac{1}{3^9} \) clearly. Calculate \( 3^9 \) carefully.

Question 6. Find the nth term of the series : 1, 2, 4, 8, ……
Answer:
Given series: 1, 2, 4, 8, ……
Now, \( \frac{2}{1} = 2, \frac{4}{2} = 2, \frac{8}{4} = 2 \)
Since \( \frac{2}{1} = \frac{4}{2} = \frac{8}{4} = \ldots = 2 \), the given sequence is a G.P.
with first term, a = 1 and common ratio, r = 2.
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow t_n = 1 \times 2^{n-1} = 2^{n-1} \)
In simple words: This is a simple G.P. where each term is double the previous term. The nth term is \( 2^{n-1} \).

📝 Teacher's Note: This is the simplest G.P. pattern. Students should memorize this: 1, 2, 4, 8, 16, 32... has nth term = \( 2^{n-1} \).

🎯 Exam Tip: For general nth term questions, always write \( t_n = ar^{n-1} \) and then substitute the values. Don't skip steps.

Question 7. Find the next three terms of the sequence : \( \sqrt{5} \), 5, \( 5\sqrt{5} \), ……
Answer:
Given sequence: \( \sqrt{5} \), 5, \( 5\sqrt{5} \), ……
Now, \( \frac{5}{\sqrt{5}} = \sqrt{5}, \frac{5\sqrt{5}}{5} = \sqrt{5} \)
Since \( \frac{5}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \ldots = \sqrt{5} \), the given sequence is a G.P.
with first term, a = \( \sqrt{5} \) and common ratio, r = \( \sqrt{5} \).
Now, \( t_n = ar^{n-1} \)
∴ Next three terms:
4th term = \( \sqrt{5} \times (\sqrt{5})^3 = \sqrt{5} \times 5\sqrt{5} = 25 \)
5th term = \( \sqrt{5} \times (\sqrt{5})^4 = \sqrt{5} \times 25 = 25\sqrt{5} \)
6th term = \( \sqrt{5} \times (\sqrt{5})^5 = \sqrt{5} \times 25\sqrt{5} = 125 \)
In simple words: Each term is \( \sqrt{5} \) times the previous term. So we multiply by \( \sqrt{5} \) to get the next term.

📝 Teacher's Note: Help students see the pattern: \( \sqrt{5}, 5, 5\sqrt{5}, 25, 25\sqrt{5}, 125 \). The terms alternate between having and not having \( \sqrt{5} \).

🎯 Exam Tip: Be careful with powers of \( \sqrt{5} \). Remember \( (\sqrt{5})^2 = 5 \), \( (\sqrt{5})^3 = 5\sqrt{5} \), \( (\sqrt{5})^4 = 25 \).

Question 8. Find the sixth term of the series : \( 2^2, 2^3, 2^4 \), ………
Answer:
Given sequence: \( 2^2, 2^3, 2^4 \), ……
Now, \( \frac{2^3}{2^2} = 2, \frac{2^4}{2^3} = 2 \)
Since \( \frac{2^3}{2^2} = \frac{2^4}{2^3} = \ldots = 2 \), the given sequence is a G.P.
with first term, a = \( 2^2 = 4 \) and common ratio, r = 2.
Now, \( t_n = ar^{n-1} \)
∴ \( t_6 = 4 \times (2)^5 = 4 \times 32 = 128 \)
In simple words: The sequence is 4, 8, 16, 32, 64, 128... Each term doubles the previous one.

📝 Teacher's Note: Show students that \( 2^2 = 4, 2^3 = 8, 2^4 = 16 \). The pattern in powers becomes a simple doubling pattern.

🎯 Exam Tip: Convert powers to actual numbers first: \( 2^2 = 4, 2^3 = 8, 2^4 = 16 \). Then find the common ratio. Always show this conversion.

Question 9. Find the seventh term of the G.P. : \( \sqrt{3} + 1, 1, \frac{\sqrt{3} - 1}{2} \), ……………..
Answer:
Given G.P.: \( \sqrt{3} + 1, 1, \frac{\sqrt{3} - 1}{2} \), ……
Here,
First term, a = \( \sqrt{3} + 1 \)
Common ratio, r = \( \frac{1}{\sqrt{3} + 1} \)
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow t_7 = (\sqrt{3} + 1) \times \left(\frac{1}{\sqrt{3} + 1}\right)^6 \)
\( = \left(\frac{1}{\sqrt{3} + 1}\right)^5 \)
\( = \left(\frac{1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\right)^5 \)
\( = \left(\frac{\sqrt{3} - 1}{2}\right)^5 \)
\( = \frac{1}{32}(\sqrt{3} - 1)^5 \)
In simple words: We use the property that \( (\sqrt{3} + 1)(\sqrt{3} - 1) = 3 - 1 = 2 \) to simplify the fraction.

📝 Teacher's Note: This uses the identity \( (a + b)(a - b) = a^2 - b^2 \). Here \( (\sqrt{3} + 1)(\sqrt{3} - 1) = 3 - 1 = 2 \).

🎯 Exam Tip: When you see \( \sqrt{3} + 1 \) and \( \sqrt{3} - 1 \), remember they are conjugates. Their product is always 2. Use this to simplify fractions.

Question 10. Find the G.P. whose first term is 64 and next term is 32.
Answer:
Given:
First term, \( a = 64 \)
Second term, \( t_2 = 32 \)

Step 1: Find the common ratio.
\( ar = 32 \)
\( \Rightarrow 64 \times r = 32 \)
\( \Rightarrow r = \frac{32}{64} = \frac{1}{2} \)

Step 2: Write the G.P.
Required G.P. = \( a, ar, ar^2, ar^3, \ldots \)
\( = 64, 32, 64 \times \left(\frac{1}{2}\right)^2, 64 \times \left(\frac{1}{2}\right)^3, \ldots \)
\( = 64, 32, 16, 8, \ldots \)

Therefore, the G.P. is 64, 32, 16, 8, ...
In simple words: We found that each term is half of the previous term. So we get 64, then 32, then 16, then 8, and so on.

📝 Teacher's Note: Show students how to find the common ratio by dividing the second term by the first term. This is the key step in G.P. problems.

🎯 Exam Tip: Always write "Given" first, then find r, then write the complete G.P. sequence. Show at least 4 terms to get full marks.

Question 11. Find the next three terms of the series: \( \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots \)
Answer:
Given sequence: \( \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots \)

Step 1: Check if it's a G.P.
\( \frac{2/9}{2/27} = 3, \frac{2/3}{2/9} = 3 \)

Since \( \frac{2/9}{2/27} = \frac{2/3}{2/9} = \ldots = 3 \), the given sequence is a G.P.
with first term, \( a = \frac{2}{27} \) and common ratio, \( r = 3 \).

Step 2: Find next three terms using \( t_n = ar^{n-1} \)
4th term = \( \frac{2}{27} \times (3)^3 = \frac{2}{27} \times 27 = 2 \)
5th term = \( \frac{2}{27} \times (3)^4 = \frac{2}{27} \times 81 = 6 \)
6th term = \( \frac{2}{27} \times (3)^5 = \frac{2}{27} \times 243 = 18 \)

Next three terms: 2, 6, 18
In simple words: Each term is 3 times the previous term. So after 2/3, we get 2, then 6, then 18.

📝 Teacher's Note: Teach students to always check the ratio between consecutive terms first. If all ratios are the same, it's a G.P.

🎯 Exam Tip: Write "Given sequence is a G.P." and show the common ratio calculation. Then use the formula to find the required terms.

Question 12. Find the next two terms of the series 2 - 6 + 18 - 54 ...
Answer:
Given series: 2 - 6 + 18 - 54 ...

Step 1: Check if it's a G.P.
\( \frac{-6}{2} = -3, \frac{18}{-6} = -3, \frac{-54}{18} = -3 \)

Since \( \frac{-6}{2} = \frac{18}{-6} = \frac{-54}{18} = \ldots = -3 \), the given sequence is a G.P.
with first term, \( a = 2 \) and common ratio, \( r = -3 \).

Step 2: Find next two terms using \( t_n = ar^{n-1} \)
5th term = \( 2 \times (-3)^4 = 2 \times 81 = 162 \)
6th term = \( 2 \times (-3)^5 = 2 \times (-243) = -486 \)

Next two terms: 162, -486
In simple words: Each term is -3 times the previous term. The signs keep changing: positive, negative, positive, negative.

📝 Teacher's Note: Point out that negative common ratio makes the signs alternate. Students often miss this pattern.

🎯 Exam Tip: Be careful with negative signs when the common ratio is negative. Show each calculation step clearly.

Exercise 11B

Question 1. Which term of the G.P.: \( -10, \frac{5}{\sqrt{3}}, \frac{5}{6}, \ldots \) is \( -\frac{5}{72} \)?
Answer:
For the given G.P.:
First term, \( a = -10 \)

Common ratio, \( r = \frac{5/\sqrt{3}}{-10} = -\frac{1}{2\sqrt{3}} \)

Step 1: Set up the equation.
If \( -\frac{5}{72} \) is the nth term of the given G.P., then
\( -\frac{5}{72} = ar^{n-1} \)

\( \Rightarrow -\frac{5}{72} = -10 \times \left(\frac{1}{2\sqrt{3}}\right)^{n-1} \)

\( \Rightarrow \frac{1}{144} = \left(\frac{1}{2\sqrt{3}}\right)^{n-1} \)

Step 2: Solve for n.
\( \Rightarrow \frac{1}{2 \times 2 \times 2 \times 2 \times \sqrt{3} \times \sqrt{3} \times \sqrt{3} \times \sqrt{3}} = \left(\frac{1}{2\sqrt{3}}\right)^{n-1} \)

\( \Rightarrow \left(\frac{1}{2\sqrt{3}}\right)^4 = \left(\frac{1}{2\sqrt{3}}\right)^{n-1} \)

\( \Rightarrow n - 1 = 4 \)
\( \Rightarrow n = 5 \)

Therefore, \( -\frac{5}{72} \) is the 5th term.
In simple words: We used the G.P. formula and found that this value appears at position 5 in the sequence.

📝 Teacher's Note: Show students how to work with fractions and surds step by step. Break down complex calculations into smaller parts.

🎯 Exam Tip: Always write the G.P. formula first, substitute values, then solve step by step. Don't skip steps in calculations.

Question 2. The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Given:
5th term = 81 \( \Rightarrow ar^4 = 81 \)
2nd term = 24 \( \Rightarrow ar = 24 \)

Step 1: Find r.
\( \frac{ar^4}{ar} = \frac{81}{24} \)
\( \Rightarrow r^3 = \frac{27}{8} \)
\( \Rightarrow r = \frac{3}{2} \)

Step 2: Find a.
\( ar = 24 \)
\( \Rightarrow a = 16 \)

Step 3: Write the G.P.
G.P. = \( a, ar, ar^2, ar^3, \ldots \)
\( = 16, 24, 16 \times \left(\frac{3}{2}\right)^2, 16 \times \left(\frac{3}{2}\right)^3, \ldots \)
\( = 16, 24, 36, 54, \ldots \)

Therefore, the G.P. is 16, 24, 36, 54, ...
In simple words: We used the two given terms to find the first term and common ratio, then wrote the complete sequence.

📝 Teacher's Note: Teach students to divide terms to eliminate 'a' and find 'r' first. This is a standard technique for G.P. problems.

🎯 Exam Tip: Always verify your answer by checking if the 2nd and 5th terms match the given values. This catches calculation errors.

Question 3. Fourth and seventh terms of a G.P. are \( \frac{1}{18} \) and \( -\frac{1}{486} \) respectively. Find the GP.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Given:
4th term = \( \frac{1}{18} \Rightarrow ar^3 = \frac{1}{18} \)
7th term = \( -\frac{1}{486} \Rightarrow ar^6 = -\frac{1}{486} \)

Step 1: Find r.
\( \frac{ar^6}{ar^3} = \frac{-1/486}{1/18} \)
\( \Rightarrow r^3 = -\frac{1}{27} \)
\( \Rightarrow r = -\frac{1}{3} \)

Step 2: Find a.
\( ar^3 = \frac{1}{18} \)
\( \Rightarrow a \times \left(-\frac{1}{3}\right)^3 = \frac{1}{18} \)
\( \Rightarrow a = -\frac{27}{18} = -\frac{3}{2} \)

Step 3: Write the G.P.
G.P. = \( a, ar, ar^2, ar^3, \ldots \)
\( = -\frac{3}{2}, -\frac{3}{2} \times \left(-\frac{1}{3}\right), -\frac{3}{2} \times \left(-\frac{1}{3}\right)^2, \frac{1}{18}, \ldots \)
\( = -\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, \ldots \)

Therefore, the G.P. is \( -\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, \ldots \)
In simple words: The common ratio is negative, so the signs keep changing. Each term is -1/3 times the previous term.

📝 Teacher's Note: When the common ratio is negative, remind students that signs alternate. Show this pattern clearly in the sequence.

🎯 Exam Tip: Be extra careful with negative fractions. Show each calculation step and check the signs at each step.

Question 4. If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Given:
1st term = \( a = 2 \)
3rd term = 8 \( \Rightarrow ar^2 = 8 \)

Step 1: Find r.
\( \frac{ar^2}{a} = \frac{8}{2} \)
\( \Rightarrow r^2 = 4 \)
\( \Rightarrow r = \pm 2 \)

Step 2: Find the second term for both values of r.
When \( a = 2 \) and \( r = 2 \)
2nd term = \( ar = 2 \times 2 = 4 \)

When \( a = 2 \) and \( r = -2 \)
2nd term = \( ar = 2 \times (-2) = -4 \)

Therefore, the second term is either 4 or -4.
In simple words: Since r² = 4, r can be +2 or -2. This gives us two possible values for the second term.

📝 Teacher's Note: When solving r² = constant, always consider both positive and negative roots. Both answers are mathematically correct.

🎯 Exam Tip: Always write both possible answers when r² gives you a perfect square. Don't forget the negative root.

Question 5. The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Given:
\( t_3 \times t_8 = 243 \)
\( \Rightarrow ar^2 \times ar^7 = 243 \)
\( \Rightarrow a^2r^9 = 243 \) ... (i)

Also, \( t_4 = 3 \)
\( \Rightarrow ar^3 = 3 \)
\( \Rightarrow a = \frac{3}{r^3} \)

Step 1: Substitute value of a in equation (i).
\( \left(\frac{3}{r^3}\right)^2 \times r^9 = 243 \)
\( \Rightarrow \frac{9}{r^6} \times r^9 = 243 \)
\( \Rightarrow 9r^3 = 243 \)
\( \Rightarrow r^3 = 27 \)
\( \Rightarrow r = 3 \)

Step 2: Find the 7th term.
\( t_7 = ar^6 = ar^3 \times r^3 = 3 \times 27 = 81 \)

Therefore, the 7th term is 81.
In simple words: We used the product of two terms and one given term to find the common ratio, then calculated the 7th term.

📝 Teacher's Note: Show students how to use the property that t₃ × t₈ = t₄ × t₇ in a G.P. This makes some calculations easier.

🎯 Exam Tip: Use the relationship between terms in G.P. to simplify calculations. Always substitute known values step by step.

Question 6. Find the geometric progression with 4th term = 54 and 7th term = 1458.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Step 1: Write the given terms using G.P. formula.
4th term = 54
\( \implies ar^3 = 54 \)
7th term = 1458
\( \implies ar^6 = 1458 \)

Step 2: Find the common ratio.
Now, \( \frac{ar^6}{ar^3} = \frac{1458}{54} \)
\( \implies r^3 = 27 \)
\( \implies r = 3 \)

Step 3: Find the first term.
\( ar^3 = 54 \)
\( \implies a \times (3)^3 = 54 \)
\( \implies a = \frac{54}{27} = 2 \)

Step 4: Write the G.P.
\( \therefore \) G.P. = a, ar, \( ar^2 \), \( ar^3 \), ...
= 2, 2 × 3, 2 × \( (3)^2 \), 54, ...
= 2, 6, 18, 54, ...

In simple words: We used the fact that in a G.P., each term is the previous term multiplied by a fixed number (common ratio). We found this ratio by dividing the 7th term by 4th term.

📝 Teacher's Note: Show students how dividing two terms gives us the common ratio raised to the power of difference in their positions. This is a key trick in G.P. problems.

🎯 Exam Tip: Always write the formula first. Then substitute values step by step. Write the final G.P. clearly with at least 4 terms to show the pattern.

Question 7. Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Step 1: Write the given information.
Now, 2nd term = \( t_2 = 6 \)
\( \implies ar = 6 \)
Also, \( t_5 = 9 \times t_3 \)
\( \implies ar^4 = 9 \times ar^2 \)
\( \implies r^2 = 9 \)
\( \implies r = ±3 \)

Step 2: Choose the positive value.
Since each term of a G.P. is positive, we have r = 3
ar = 6
\( \implies a \times 3 = 6 \)
\( \implies a = 2 \)

Step 3: Write the G.P.
\( \therefore \) G.P. = a, ar, \( ar^2 \), \( ar^3 \), ...
= 2, 6, 2 × \( (3)^2 \), 2 × \( (3)^3 \), ...
= 2, 6, 18, 54, ...

In simple words: We found that the 5th term is 9 times the 3rd term. This helped us find the common ratio. Since all terms must be positive, we picked the positive value of the ratio.

📝 Teacher's Note: Remind students that when we take square roots, we get both positive and negative values. The condition "all terms positive" helps us choose the right one.

🎯 Exam Tip: When you get r = ±3, always check the given conditions to pick the right sign. Write "Since all terms are positive" to show your reasoning.

Question 8. The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Step 1: Write the given terms.
Now,
4th term = \( t_4 = 10 \)
\( \implies ar^3 = 10 \)
7th term = \( t_7 = 80 \)
\( \implies ar^6 = 80 \)

Step 2: Find the common ratio.
\( \frac{ar^6}{ar^3} = \frac{80}{10} \)
\( \implies r^3 = 8 \)
\( \implies r = 2 \)

Step 3: Find the first term.
\( ar^3 = 10 \)
\( \implies a \times (2)^3 = 10 \)
\( \implies a = \frac{10}{8} = \frac{5}{4} \)

Step 4: Find the number of terms.
Last term = l = 2560
Let there be n terms in given G.P.
\( \implies t_n = 2560 \)
\( \implies ar^{n-1} = 2560 \)
\( \implies \frac{5}{4} \times (2)^{n-1} = 2560 \)
\( \implies (2)^{n-1} = 2048 \)
\( \implies (2)^{n-1} = (2)^{11} \)
\( \implies n - 1 = 11 \)
\( \implies n = 12 \)

Thus, we have
First term = \( \frac{5}{4} \), Common ratio = 2 and Number of terms = 12

In simple words: We found the ratio by dividing two given terms. Then we found the first term. Finally, we used the last term to find how many terms are in the sequence.

📝 Teacher's Note: Show students how 2048 = 2¹¹ by repeated doubling: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048. This helps them recognize powers of 2.

🎯 Exam Tip: Always check your answer by verifying one of the given terms. For example, check if the 7th term really equals 80 using your values.

Question 9. If the 4th and 9th terms of a G.P. are 54 and 13122 respectively, find the GP. Also, find its general term.
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Step 1: Write the given terms.
Now,
4th term = \( t_4 = 54 \)
\( \implies ar^3 = 54 \)
9th term = \( t_9 = 13122 \)
\( \implies ar^8 = 13122 \)

Step 2: Find the common ratio.
\( \frac{ar^8}{ar^3} = \frac{13122}{54} \)
\( \implies r^5 = 243 \)
\( \implies r = 3 \)

Step 3: Find the first term.
\( ar^3 = 54 \)
\( \implies a \times (3)^3 = 54 \)
\( \implies a = \frac{54}{27} = 2 \)

Step 4: Write the G.P. and general term.
\( \therefore \) Required G.P. = a, ar, \( ar^2 \), \( ar^3 \), ...
= 2, 2 × 3, 2 × \( (3)^2 \), 54
= 2, 6, 18, 54

General term = \( t_n = ar^{n-1} = 2 \times (3)^{n-1} \)

In simple words: We found the common ratio by dividing the 9th term by 4th term. Then we found the first term. The general term gives us a formula to find any term in the sequence.

📝 Teacher's Note: Help students see that 243 = 3⁵ by showing: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243. This pattern helps in exams.

🎯 Exam Tip: The general term formula \( t_n = ar^{n-1} \) is very important. Always write it clearly. You can use it to find any term without writing the whole sequence.

Question 10. The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : \( q^2 = pr \).
Answer:
Let the first term of the G.P. be a and its common ratio be r.

Step 1: Write the given terms using G.P. formula.
5th term = \( t_5 = p \)
\( \implies ar^4 = p \)

8th term = \( t_8 = q \)
\( \implies ar^7 = q \)

11th term = \( t_{11} = r \)
\( \implies ar^{10} = r \)

Step 2: Find the product pr.
Now,
pr = \( ar^4 \times ar^{10} = a^2 \times r^{14} = (a \times r^7)^2 = q^2 \)
\( \implies q^2 = pr \)

In simple words: We used the rule that when we multiply two terms in a G.P., we add their powers. The middle term squared equals the product of the other two terms.

📝 Teacher's Note: This is a special property of G.P. If three terms are equally spaced (like 5th, 8th, 11th), the middle term squared equals the product of the outer terms.

🎯 Exam Tip: This property \( q^2 = pr \) works for any three equally spaced terms in a G.P. Remember this rule — it saves time in exams.

Exercise 11C

Question 1. Find the seventh term from the end of the series : \( \sqrt{2} \), 2, \( 2\sqrt{2} \), ......... 32.
Answer:
Given series: \( \sqrt{2} \), 2, \( 2\sqrt{2} \), ..., 32

Step 1: Find the common ratio.
Now, \( \frac{2}{\sqrt{2}} = \sqrt{2} \), \( \frac{2\sqrt{2}}{2} = \sqrt{2} \)

So, the given series is a G.P. with common ratio, r = \( \sqrt{2} \)
Here, last term, l = 32

Step 2: Find the 7th term from the end.
\( \therefore \) 7th term from an end = \( \frac{l}{r^6} = \frac{32}{(\sqrt{2})^6} = \frac{32}{8} = 4 \)

In simple words: To find terms from the end, we divide the last term by the common ratio raised to the appropriate power. This is like going backwards in the sequence.

📝 Teacher's Note: Show students that \( (\sqrt{2})^6 = (\sqrt{2})^6 = (2^{1/2})^6 = 2^3 = 8 \). This step-by-step calculation helps avoid mistakes.

🎯 Exam Tip: For nth term from the end, use the formula \( \frac{l}{r^{n-1}} \) where l is the last term. Don't forget to subtract 1 from n.

Question 2. Find the third term from the end of the GP. \( \frac{2}{27} \), \( \frac{2}{9} \), \( \frac{2}{3} \), ............. 162
Answer:
Given G.P.: \( \frac{2}{27} \), \( \frac{2}{9} \), \( \frac{2}{3} \), ....., 162

Step 1: Find the common ratio.
Here,
Common ratio, r = \( \frac{2/9}{2/27} = 3 \)

Last term, l = 162

Step 2: Find the 3rd term from the end.
\( \therefore \) 3rd term from an end = \( \frac{l}{r^2} = \frac{162}{(3)^2} = \frac{162}{9} = 18 \)

In simple words: We found the common ratio by dividing any term by the previous term. Then we used the formula to find the 3rd term from the end by going backwards.

📝 Teacher's Note: When dividing fractions, remind students to multiply by the reciprocal: \( \frac{2/9}{2/27} = \frac{2}{9} \times \frac{27}{2} = 3 \).

🎯 Exam Tip: Always verify your common ratio by checking with another pair of consecutive terms. This prevents calculation errors.

Question 3. For the \( \frac{1}{27} \), \( \frac{1}{9} \), \( \frac{1}{3} \), ........... 81; find the product of fourth term from the beginning and the fourth term from the end.
Answer:
Given G.P.: \( \frac{1}{27} \), \( \frac{1}{9} \), \( \frac{1}{3} \), ....., 81

Step 1: Find the common ratio and terms.
Here,
Common ratio, r = \( \frac{1/9}{1/27} = 3 \)

First term, a = \( \frac{1}{27} \) and Last term, l = 81

Step 2: Find the required terms.
\( \therefore \) 4th term from the beginning = \( ar^3 = \frac{1}{27} \times (3)^3 = \frac{1}{27} \times 27 = 1 \)

And, 4th term from an end = \( \frac{l}{r^3} = \frac{81}{(3)^3} = \frac{81}{27} = 3 \)

Step 3: Find the product.
Thus, required product = 1 × 3 = 3

In simple words: We found the 4th term from the start using the forward formula and the 4th term from the end using the backward formula. Then we multiplied them.

📝 Teacher's Note: Show students that in a G.P., terms equidistant from the ends have a special relationship. Their product often gives a simple answer.

🎯 Exam Tip: For terms from the beginning, use \( ar^{n-1} \). For terms from the end, use \( \frac{l}{r^{n-1}} \). Keep these two formulas separate to avoid confusion.

Question 4. If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that: (q - r) log a + (r - p) log b + (p - q) log c = 0
Answer: Let the first term of the G.P. be A and its common ratio be R.
Then,
pth term = a
\( \implies AR^{p-1} = a \)
qth term = b
\( \implies AR^{q-1} = b \)
rth term = c
\( \implies AR^{r-1} = c \)
Now,
\( a^{q-r} \times b^{r-p} \times c^{p-q} = (AR^{p-1})^{q-r} \times (AR^{q-1})^{r-p} \times (AR^{r-1})^{p-q} \)
\( = A^{q-r} \cdot R^{(p-1)(q-r)} \times A^{r-p} \cdot R^{(q-1)(r-p)} \times A^{p-q} \cdot R^{(r-1)(p-q)} \)
\( = A^{q-r+r-p+p-q} \cdot R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \)
\( = A^0 \times R^0 \)
\( = 1 \)
Taking log on both the sides, we get
\( \log(a^{q-r} \times b^{r-p} \times c^{p-q}) = \log 1 \)
\( \implies (q-r)\log a + (r-p)\log b + (p-q)\log c = 0 \) .....(proved)
In simple words: We use the general form of G.P. terms to show that when we multiply the given powers and take log, everything cancels out to give zero. This is a special property of G.P.

📝 Teacher's Note: Help students see the pattern - the exponents are differences like (q-r), (r-p), (p-q). When added, they equal zero. This cancellation is key to the proof.

🎯 Exam Tip: Always write the general term first, then substitute the given values. Show clearly that the product equals 1 before taking logarithm. Write "proved" at the end.

Question 5. If a, b and c in G.P., prove that: log an, log bn and log cn are in A.P.
Answer: Here, a, b, c are in G.P.
\( \implies b^2 = ac \)
Taking log on both sides, we get
\( \log(b^2) = \log(ac) \)
\( \implies 2\log b = \log a + \log c \)
\( \implies \log b + \log b = \log a + \log c \)
\( \implies \log b - \log a = \log c - \log b \)
\( \implies \log a, \log b \text{ and } \log c \text{ are in A.P.} \)
Now multiplying each term by n:
\( n\log a, n\log b, n\log c \text{ are in A.P.} \)
\( \implies \log a^n, \log b^n, \log c^n \text{ are in A.P.} \)
In simple words: When numbers are in G.P., their logs are in A.P. If we multiply each log by the same number n, they stay in A.P.

📝 Teacher's Note: Show students that log converts multiplication (G.P. property) to addition (A.P. property). This is why logarithms are so useful in mathematics.

🎯 Exam Tip: Start with the G.P. condition b² = ac, then take log of both sides. Use the property that log(xy) = log x + log y. Write clearly that multiplying by n keeps the A.P. property.

Question 6. If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Answer: Let a₁, a₂, a₃, ......, aₙ, ..... be a G.P. with common ratio r.
\( \implies \frac{a_{n+1}}{a_n} = r \text{ for all } n \in N \)
If each term of a G.P. is raised to the power x, we get the sequence
\( a_1^x, a_2^x, a_3^x, ....., a_n^x, ........ \)
Now, \( \frac{(a_{n+1})^x}{(a_n)^x} = \left(\frac{a_{n+1}}{a_n}\right)^x = r^x \text{ for all } n \in N \)
Hence, \( a_1^x, a_2^x, a_3^x, ....., a_n^x, ..... \text{ is also a G.P.} \)
In simple words: When you raise all terms of a G.P. to the same power, the ratio between consecutive terms stays constant. So it remains a G.P.

📝 Teacher's Note: Use a simple example like 2, 4, 8, 16 (ratio = 2). If we square each: 4, 16, 64, 256 (ratio = 4 = 2²). The pattern holds.

🎯 Exam Tip: Show the ratio formula clearly. Use the power rule (a/b)ˣ = aˣ/bˣ. Conclude that the new common ratio is rˣ where r was the original ratio.

Question 7. If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that: x², b², y² are in A.P.
Answer: a, b and c are in A.P.
\( \implies 2b = a + c \)
a, x and b are in G.P.
\( \implies x^2 = ab \)
b, y and c are in G.P.
\( \implies y^2 = bc \)
Now,
\( x^2 + y^2 = ab + bc \)
\( = b(a + c) \)
\( = b \times 2b \)
\( = 2b^2 \)
\( \implies x^2, b^2 \text{ and } y^2 \text{ are in A.P.} \)
In simple words: We use the G.P. property (middle term squared = product of outer terms) and A.P. property (middle term doubled = sum of outer terms) to show the required result.

📝 Teacher's Note: Emphasize that we use both A.P. and G.P. conditions together. The key step is substituting 2b for (a + c) from the A.P. condition.

🎯 Exam Tip: Write all given conditions first. Use x² = ab and y² = bc from G.P., then substitute a + c = 2b from A.P. Show that x² + y² = 2b².

Question 8. If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that:
(i) \( \frac{1}{x} + \frac{1}{y} = \frac{2}{b} \)
(ii) \( \frac{a}{x} + \frac{c}{y} = 2 \)
Answer:

Solution 8(i).
a, b and c are in G.P.
\( \implies b^2 = ac \)
a, x, b, y and c are in A.P.
\( \implies 2x = a + b \implies x = \frac{a + b}{2} \)
\( 2b = x + y \implies b = \frac{x + y}{2} \)
\( 2y = b + c \implies y = \frac{b + c}{2} \)
Now,
\( \frac{1}{x} + \frac{1}{y} = \frac{2}{a + b} + \frac{2}{b + c} \)
\( = \frac{2b + 2c + 2a + 2b}{ab + ac + b^2 + bc} \)
\( = \frac{2a + 2c + 4b}{ab + b^2 + b^2 + bc} \)
\( = \frac{2a + 2c + 4b}{ab + 2b^2 + bc} \)
\( = \frac{2(a + c + 2b)}{b(a + 2b + c)} \)
\( = \frac{2}{b} \)
In simple words: We find x and y in terms of a, b, c using A.P. property, then substitute to prove the given result.

Solution 8(ii).
a, b and c are in G.P.
\( \implies b^2 = ac \)
a, x, b, y and c are in A.P.
\( \implies 2x = a + b \implies x = \frac{a + b}{2} \)
\( 2b = x + y \implies b = \frac{x + y}{2} \)
\( 2y = b + c \implies y = \frac{b + c}{2} \)
Now,
\( \frac{a}{x} + \frac{c}{y} = \frac{2a}{a + b} + \frac{2c}{b + c} \)
\( = \frac{2a(b + c) + 2c(a + b)}{(a + b)(b + c)} \)
\( = \frac{2ab + 2ac + 2ac + 2bc}{ab + ac + b^2 + bc} \)
\( = \frac{2ab + 4ac + 2bc}{ab + b^2 + b^2 + bc} \)
\( = \frac{2(ab + 2ac + bc)}{ab + 2b^2 + bc} \)
\( = \frac{2(ab + 2ac + bc)}{ab + 2ac + bc} \)
\( = 2 \)
In simple words: Similar to part (i), we substitute the A.P. relations and use the G.P. condition b² = ac to simplify and get the answer 2.

📝 Teacher's Note: This problem combines A.P. and G.P. properties. Show students how to express x and y first, then substitute carefully. The algebra needs patience.

🎯 Exam Tip: Write the A.P. condition as 2x = a + b, not x = (a + b)/2 initially. This makes the algebra cleaner. Use b² = ac at the right moment to simplify.

Question 9. If a, b and c are in A.P. and also in G.P., show that: a = b = c.
Answer: a, b and c are in A.P.
\( \implies 2b = a + c \)
\( \implies b = \frac{a + c}{2} \)
a, b and c are also in G.P.
\( \implies b^2 = ac \)
\( \implies \left(\frac{a + c}{2}\right)^2 = ac \)
\( \implies \frac{a^2 + c^2 + 2ac}{4} = ac \)
\( \implies a^2 + c^2 + 2ac = 4ac \)
\( \implies a^2 + c^2 - 2ac = 0 \)
\( \implies (a - c)^2 = 0 \)
\( \implies a - c = 0 \)
\( \implies a = c \)
Now, 2b = a + c
\( \implies 2b = a + a \)
\( \implies 2b = 2a \)
\( \implies b = a \)
Thus, we have a = b = c
In simple words: When three numbers are both in A.P. and G.P., they must all be equal. This is because the conditions force them to be the same.

📝 Teacher's Note: This shows that constant sequences are the only sequences that are both A.P. and G.P. Give examples like 3, 3, 3 or 7, 7, 7.

🎯 Exam Tip: Substitute b = (a + c)/2 from A.P. into b² = ac from G.P. Expand carefully to get (a - c)² = 0. This immediately gives a = c.

Question 10. The first term of a G.P. is a and its nth term is b, where n is an even number. If the product of first n numbers of this G.P. is P; prove that: p² = (ab)ⁿ.
Answer: For a G.P.,
First term = a
Let the common ratio = r
nth term = b
\( \implies ar^{n-1} = b \)
P = Product of first n numbers of the given G.P.
\( \implies P = a \times ar \times ar^2 \times ar^3 \times ....... \times ar^{n-1} \)
\( \implies P = a \times ar \times ar^2 \times ar^3 \times ....... \times b \)
\( \implies P = a \times ar \times ar^2 \times ar^3 \times ....... \times \frac{b}{r^2} \times \frac{b}{r} \times b \)
\( \implies P = (ab) \times \left(ar \times \frac{b}{r}\right) \times \left(ar^2 \times \frac{b}{r^2}\right) \times ....... \frac{n}{2} \text{ terms} \)
\( \implies P = (ab) \times (ab) \times (ab) \times ...... \frac{n}{2} \text{ terms} \)
\( \implies P = (ab)^{\frac{n}{2}} \)
\( \implies P = \sqrt{ab^n} \)
\( \implies P^2 = ab^n \)
In simple words: We pair the terms from both ends. Each pair multiplies to give ab. Since there are n/2 pairs, the product is (ab)^(n/2).

📝 Teacher's Note: The key insight is pairing terms: first with last, second with second-last, etc. Since n is even, we get n/2 pairs, each giving product ab.

🎯 Exam Tip: Write the product clearly showing all terms. Use the fact that n is even to pair terms symmetrically. Show that each pair gives ab.

Question 11. If a, b, c and d are consecutive terms of a G.P.; prove that: (a² + b²), (b² + c²) and (c² + d²) are in GP.
Answer: Since a, b, c, d are consecutive terms of a G.P., let the common ratio be r.
Then: b = ar, c = ar², d = ar³
Now we need to prove that (a² + b²), (b² + c²) and (c² + d²) are in G.P.
For this, we need to show: (b² + c²)² = (a² + b²)(c² + d²)

LHS = (b² + c²)²
\( = (a²r² + a²r⁴)² \)
\( = a⁴(r² + r⁴)² \)
\( = a⁴r⁴(1 + r²)² \)

RHS = (a² + b²)(c² + d²)
\( = (a² + a²r²)(a²r⁴ + a²r⁶) \)
\( = a²(1 + r²) \times a²r⁴(1 + r²) \)
\( = a⁴r⁴(1 + r²)² \)

Since LHS = RHS, we have (b² + c²)² = (a² + b²)(c² + d²)
Therefore, (a² + b²), (b² + c²) and (c² + d²) are in G.P.
In simple words: We express all terms using the common ratio r, then check if the middle term squared equals the product of the other two terms.

📝 Teacher's Note: Show students that we check the G.P. condition by verifying (middle term)² = (first term) × (third term). Factor out common terms to simplify.

🎯 Exam Tip: Express all terms using the first term and common ratio. Calculate both sides of the G.P. condition separately, then show they are equal. Factor carefully to avoid errors.

Question 12. If a, b, c and d are consecutive terms of a G.P. To prove: \( \frac{1}{a^2 + b^2} \), \( \frac{1}{b^2 + c^2} \) and \( \frac{1}{c^2 + d^2} \) are in G.P.
Answer: Let r be the common ratio of this G.P.
Given: a, b, c, d are in G.P.
\( \Rightarrow 1^{st} = a \)
\( 2^{nd} \) term = b = ar
\( 3^{rd} \) term = c = \( ar^2 \)
\( 4^{th} \) term = d = \( ar^3 \)

Now, \( (b^2 + c^2)^2 = [(ar)^2 + (ar^2)^2]^2 \)
\( = [a^2r^2 + a^2r^4]^2 \)
\( = [a^2r^2(1 + r^2)]^2 \)
\( = a^4r^4(1 + r^2)^2 \)

And, \( (a^2 + b^2) \times (c^2 + d^2) = [a^2 + (ar)^2] \times [(ar^2)^2 + (ar^3)^2] \)
\( = [a^2 + a^2r^2] \times [a^2r^4 + a^2r^6] \)
\( = a^2(1 + r^2) \times a^2r^4(1 + r^2) \)
\( = a^4r^4(1 + r^2)^2 \)

\( \Rightarrow (b^2 + c^2)^2 = (a^2 + b^2) \times (c^2 + d^2) \)
i.e. \( \frac{b^2 + c^2}{a^2 + b^2} = \frac{c^2 + d^2}{b^2 + c^2} \)

Hence, \( (a^2 + b^2) \), \( (b^2 + c^2) \) and \( (c^2 + d^2) \) are in G.P.

Therefore, \( \frac{1}{a^2 + b^2} \), \( \frac{1}{b^2 + c^2} \) and \( \frac{1}{c^2 + d^2} \) are in G.P.
In simple words: If we prove that three numbers are in G.P., then their reciprocals are also in G.P. We showed that the denominators form a G.P., so their reciprocals also form a G.P.

📝 Teacher's Note: Use the property that if \( p, q, r \) are in G.P., then \( q^2 = pr \). Students should remember that reciprocals of terms in G.P. also form G.P.

🎯 Exam Tip: Write clearly that \( (b^2 + c^2)^2 = (a^2 + b^2)(c^2 + d^2) \). This is the key step. Show all algebraic working step by step.

Exercise 11D

Question 1. Find the sum of G.P.:
(i) 1 + 3 + 9 + 27 + ... to 12 terms.
(ii) 0.3 + 0.03 + 0.003 + 0.0003 + ... to 8 terms.
(iii) \( 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \)... to 9 terms.
(iv) \( 1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \)... to n terms.
(v) \( \frac{x+y}{x-y} + 1 + \frac{x-y}{x+y} + \)... upto n terms.
(vi) \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \)... to n terms.

📝 Teacher's Note: First identify the first term 'a' and common ratio 'r' for each series. Then use the formula \( S_n = \frac{a(r^n - 1)}{r - 1} \) when r > 1 and \( S_n = \frac{a(1 - r^n)}{1 - r} \) when r < 1.

🎯 Exam Tip: Always write the formula you are using. Show all substitutions clearly. Check if r > 1 or r < 1 to use the correct formula.

Solution 1(i).
Answer: Given G.P.: 1 + 3 + 9 + 27 + ...
Here, first term, a = 1
common ratio, r = \( \frac{3}{1} = 3 \) (r > 1)
number of terms to be added, n = 12

\( S_n = \frac{a(r^n - 1)}{r - 1} \)

\( S_{12} = \frac{1(3^{12} - 1)}{3 - 1} = \frac{3^{12} - 1}{2} = \frac{531441 - 1}{2} = \frac{531440}{2} = 265720 \)
In simple words: We use the sum formula for G.P. when the common ratio is greater than 1. We substitute a = 1, r = 3, n = 12 and calculate the sum.

📝 Teacher's Note: Show students how \( 3^{12} = 531441 \) step by step. Common mistake is using wrong formula when r > 1.

🎯 Exam Tip: Write "Given", identify a and r clearly. Use the correct formula based on whether r > 1 or r < 1. Show the final numerical answer.

Solution 1(ii).
Answer: Given G.P.: 0.3 + 0.03 + 0.003 + 0.0003 + ...
Here, first term, a = 0.3
common ratio, r = \( \frac{0.03}{0.3} = 0.1 \) (r < 1)
number of terms to be added, n = 8

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_8 = \frac{0.3(1 - (0.1)^8)}{1 - 0.1} = \frac{0.3(1 - (0.1)^8)}{0.9} = \frac{1 - (0.1)^8}{3} = \frac{1}{3}(1 - \frac{1}{10^8}) \)
In simple words: Each term is one-tenth of the previous term. Since r < 1, we use the formula for decreasing G.P. The sum comes out to be approximately \( \frac{1}{3} \).

📝 Teacher's Note: Help students see that 0.03/0.3 = 1/10 = 0.1. Since (0.1)^8 is very small, the answer is very close to 1/3.

🎯 Exam Tip: Always check if r < 1 or r > 1. For r < 1, use the formula with (1 - r^n). Show decimal to fraction conversion clearly.

Solution 1(iii).
Answer: Given G.P.: \( 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \)...
Here, first term, a = 1
common ratio, r = \( \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \) (r < 1)
number of terms to be added, n = 9

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_9 = \frac{1(1 - (-\frac{1}{2})^9)}{1 - (-\frac{1}{2})} = \frac{1 - (-\frac{1}{2})^9}{1 + \frac{1}{2}} = \frac{1 - (-\frac{1}{2})^9}{\frac{3}{2}} \)

\( = \frac{2}{3}(1 + \frac{1}{2^9}) = \frac{2}{3}(1 + \frac{1}{512}) = \frac{2}{3} \times \frac{513}{512} = \frac{171}{256} \)
In simple words: This is an alternating G.P. with positive and negative terms. The common ratio is negative (-1/2). We calculate the sum using the standard formula.

📝 Teacher's Note: Point out that (-1/2)^9 = -1/512 (odd power gives negative). Students often miss the sign when the ratio is negative.

🎯 Exam Tip: For negative common ratio, be very careful with signs. Write (-1/2)^9 = -1/512 clearly. Simplify the final fraction completely.

Solution 1(iv).
Answer: Given G.P.: \( 1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \)... upto n terms
Here, first term, a = 1
common ratio, r = \( \frac{-\frac{1}{3}}{1} = -\frac{1}{3} \) (r < 1)
number of terms to be added = n

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_n = \frac{1(1 - (-\frac{1}{3})^n)}{1 - (-\frac{1}{3})} = \frac{1 - (-\frac{1}{3})^n}{1 + \frac{1}{3}} = \frac{1 - (-\frac{1}{3})^n}{\frac{4}{3}} \)

\( = \frac{3}{4}[1 - (-\frac{1}{3})^n] \)
In simple words: This is an alternating series where terms get smaller. The final answer depends on whether n is even or odd because of the negative ratio.

📝 Teacher's Note: Explain that (-1/3)^n will be positive if n is even and negative if n is odd. This affects the final answer.

🎯 Exam Tip: Keep the answer in terms of n. Write the formula clearly with (-1/3)^n. Don't try to simplify further unless n is given a specific value.

Solution 1(v).
Answer:
Given G.P.: \( \frac{x + y}{x - y} + 1 + \frac{x - y}{x + y} + \ldots \) up to n terms

Here,
first term, a = \( \frac{x + y}{x - y} \)

common ratio, r = \( \frac{1}{\frac{x + y}{x - y}} = \frac{x - y}{x + y} \) (r < 1)

number of terms to be added = n

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_n = \frac{\frac{x + y}{x - y} \left[ 1 - \left( \frac{x - y}{x + y} \right)^n \right]}{1 - \frac{x - y}{x + y}} \)

\( = \frac{\frac{x + y}{x - y} \left[ 1 - \left( \frac{x - y}{x + y} \right)^n \right]}{\frac{x + y - x + y}{x + y}} \)

\( = \frac{\frac{x + y}{x - y} \left[ 1 - \left( \frac{x - y}{x + y} \right)^n \right]}{\frac{2y}{x + y}} \)

\( = \frac{(x + y)^2 \left[ 1 - \left( \frac{x - y}{x + y} \right)^n \right]}{2y(x - y)} \)

In simple words: This is a geometric series formula. We find the first term and common ratio, then use the sum formula for n terms.

📝 Teacher's Note: Help students identify the first term and common ratio step by step. Show that dividing consecutive terms gives the common ratio. Practice with simpler numbers first.

🎯 Exam Tip: Always write "Given G.P." and identify a and r clearly. Write the formula \( S_n = \frac{a(1-r^n)}{1-r} \) before substituting. Show each step of simplification.

Solution 1(vi).
Answer:
Given G.P.: \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \ldots \) up to n terms

Here,
first term, a = \( \sqrt{3} \)

common ratio, r = \( \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3} \) (r < 1)

number of terms to be added = n

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_n = \frac{\sqrt{3} \left[ 1 - \left( \frac{1}{3} \right)^n \right]}{1 - \frac{1}{3}} \)

\( = \frac{\sqrt{3} \left( 1 - \frac{1}{3^n} \right)}{\frac{2}{3}} \)

\( = \frac{3\sqrt{3}}{2} \left( 1 - \frac{1}{3^n} \right) \)

In simple words: We have a decreasing geometric series. Each term is one-third of the previous term. The sum gets closer to a fixed value as we add more terms.

📝 Teacher's Note: Show students how to find the common ratio by dividing the second term by the first term. Use simple fractions first to make the concept clear.

🎯 Exam Tip: Write the common ratio as a simple fraction. Show that r < 1 for decreasing series. Always simplify the final answer completely.

Question 2. How many terms of the geometric progression 1+4 + 16 + 64 + ……… must be added to get sum equal to 5461?
Answer:
Given G.P.: 1 + 4 + 16 + 64 + ...

Here,
first term, a = 1
common ratio, r = \( \frac{4}{1} = 4 \) (r > 1)

Let the number of terms to be added = n
Then, \( S_n = 5461 \)

\( \frac{a(r^n - 1)}{r - 1} = 5461 \)

\( \frac{1(4^n - 1)}{4 - 1} = 5461 \)

\( \frac{4^n - 1}{3} = 5461 \)

\( 4^n - 1 = 16383 \)

\( 4^n = 16384 \)

\( 4^n = 4^7 \)

\( n = 7 \)

Hence, required number of terms = 7

In simple words: This is an increasing series where each term is 4 times the previous term. We use the formula for sum and solve for n to find how many terms we need.

📝 Teacher's Note: Show students that 4^7 = 16384 by calculating step by step. Point out that for r > 1, we use the formula with (r^n - 1) in the numerator, not (1 - r^n).

🎯 Exam Tip: For r > 1, use \( S_n = \frac{a(r^n - 1)}{r - 1} \). Show all calculation steps. Check your answer by substituting back into the original equation.

Question 3. The first term of a G.P. is 27 and its 8th term is \( \frac{1}{81} \). Find the sum of its first 10 terms.
Answer:
Given,
First term, a = 27
8th term = \( ar^7 = \frac{1}{81} \)
n = 10

Now,
\( \frac{ar^7}{a} = \frac{1/81}{27} \)

\( r^7 = \frac{1}{2187} \)

\( r^7 = \left( \frac{1}{3} \right)^7 \)

\( r = \frac{1}{3} \) (r < 1)

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_{10} = \frac{27 \left[ 1 - \left( \frac{1}{3} \right)^{10} \right]}{1 - \frac{1}{3}} \)

\( = \frac{27 \left( 1 - \frac{1}{3^{10}} \right)}{\frac{2}{3}} \)

\( = \frac{81}{2} \left( 1 - \frac{1}{3^{10}} \right) \)

In simple words: We first find the common ratio using the given terms. Since the 8th term is much smaller than the first term, the series is decreasing. Each term is one-third of the previous term.

📝 Teacher's Note: Help students see that 2187 = 3^7, so 1/2187 = (1/3)^7. This is key to finding the common ratio. Show this calculation step by step.

🎯 Exam Tip: Use the nth term formula \( a_n = ar^{n-1} \) to find r first. Then use the sum formula. Show that 3^{10} is a very large number, so 1/3^{10} is very small.

Question 4. A boy spends Rs. 10 on first day, Rs. 20 on second day, Rs. 40 on third day and so on. Find how much, in all, will he spend in 12 days?
Answer:
Amount spent on 1st day = Rs. 10
Amount spent on 2nd day = Rs. 20
Amount spent on 3rd day = Rs. 40 and so on

Now, \( \frac{20}{10} = 2, \frac{40}{20} = 2 \)

Thus, 10, 20, 40, ... is a G.P. with first term, a = 10
and common ratio, r = 2 (r > 1)

Total amount spent in 12 days = \( S_{12} \)

\( S_n = \frac{a(r^n - 1)}{r - 1} \)

\( S_{12} = \frac{10(2^{12} - 1)}{2 - 1} = 10(2^{12} - 1) = 10(4096 - 1) = 10 \times 4095 = 40950 \)

Hence, the total amount spent in 12 days is Rs. 40950.

In simple words: The boy doubles his spending every day. This forms a geometric progression. We use the sum formula to find total spending over 12 days.

📝 Teacher's Note: Help students recognize the doubling pattern. Show that 2^12 = 4096 by repeated multiplication or using calculator. This is a real-life application of geometric progression.

🎯 Exam Tip: Always check if consecutive terms have a constant ratio to confirm it's a G.P. For r > 1, use \( S_n = \frac{a(r^n - 1)}{r - 1} \). Write the final answer with units (Rs.).

Question 5. The 4th and the 7th terms of a G.P. are \( \frac{1}{27} \) and \( \frac{1}{729} \) respectively. Find the sum of n terms of this G.P.
Answer:
For a G.P.,
4th term = \( ar^3 = \frac{1}{27} \)
7th term = \( ar^6 = \frac{1}{729} \)

Now, \( \frac{ar^6}{ar^3} = \frac{1/729}{1/27} \)

\( r^3 = \frac{1}{27} = \left( \frac{1}{3} \right)^3 \)

\( r = \frac{1}{3} \) (r < 1)

\( a \times \frac{1}{27} = \frac{1}{27} \)

\( a = 1 \)

\( S_n = \frac{a(1 - r^n)}{1 - r} \)

\( S_n = \frac{1 \left[ 1 - \left( \frac{1}{3} \right)^n \right]}{1 - \frac{1}{3}} \)

\( = \frac{\left( 1 - \frac{1}{3^n} \right)}{\frac{2}{3}} \)

\( = \frac{3}{2} \left( 1 - \frac{1}{3^n} \right) \)

In simple words: We use two given terms to find the common ratio and first term. Then we apply the sum formula for any number of terms n.

📝 Teacher's Note: Show students that dividing the 7th term by 4th term gives r^3. This is because the powers subtract: r^6 ÷ r^3 = r^(6-3) = r^3.

🎯 Exam Tip: To find r, divide any term by the term that comes 3 positions before it to get r^3. Then take cube root. Find a by substituting r back into any given term equation.

Question 6. A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.
Answer:
For a G.P.,
Common ratio, r = 3 (r > 1)
Last term, l = 486
S = 728

\( \frac{lr - a}{r - 1} = 728 \)

\( \frac{486 \times 3 - a}{3 - 1} = 728 \)

\( \frac{1458 - a}{2} = 728 \)

\( 1458 - a = 1456 \)

Hence, the first term is 2.

In simple words: When we know the last term and sum, we use a special formula. We solve the equation to find the first term of the progression.

📝 Teacher's Note: Teach students the formula \( S = \frac{lr - a}{r - 1} \) when the last term is known. This is different from the usual formula with r^n.

🎯 Exam Tip: For problems with last term given, use \( S = \frac{lr - a}{r - 1} \) where l is the last term. This formula is easier than finding n first and then using the regular formula.

Question 7. Find the sum of G.P.: 3, 6, 12, ……………. 1536.
Answer:
Given: G.P.: 3, 6, 12, ………, 1536
Here,
First term, a = 3
Common ratio, \( r = \frac{6}{3} = 2 \) (r > 1)
Last term, l = 1536

Required sum = \( \frac{lr - a}{r - 1} \)
= \( \frac{1536 \times 2 - 3}{2 - 1} \)
= \( \frac{3072 - 3}{1} \)
= 3069
In simple words: We used the sum formula for G.P. when we know the first term, common ratio, and last term. We multiply the last term by the ratio and subtract the first term, then divide by (ratio - 1).

📝 Teacher's Note: Show students how to identify first term and common ratio. Make them check that 6÷3 = 2 and 12÷6 = 2. This confirms it is a G.P.

🎯 Exam Tip: Always write the formula first. Check if r > 1 or r < 1 to use the right formula. Show all substitution steps clearly.

Question 8. How many terms of the series 2 + 6 + 18 + ………….. must be taken to make the sum equal to 728?
Answer:
Given series: 2 + 6 + 18 + ……
Now, \( \frac{6}{2} = 3, \frac{18}{6} = 3 \)
Thus, given series is a G.P. with first term, a = 2
and common ratio, r = 3 (r > 1)
Let the number of terms to be added = n
Then, \( S_n = 728 \)

\( \Rightarrow \frac{a(r^n - 1)}{r - 1} = 728 \)
\( \Rightarrow \frac{2(3^n - 1)}{3 - 1} = 728 \)
\( \Rightarrow 3^n - 1 = 728 \)
\( \Rightarrow 3^n = 729 \)
\( \Rightarrow 3^n = 3^6 \)
\( \Rightarrow n = 6 \)

Hence, required number of terms = 6
In simple words: We found the pattern (each term is 3 times the previous one). Then we used the sum formula and solved for n by comparing powers of 3.

📝 Teacher's Note: Make students verify that 729 = 3⁶ by calculating step by step: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81, 3⁵ = 243, 3⁶ = 729.

🎯 Exam Tip: Write "Given series is G.P. with a = ... and r = ..." clearly. Always verify your answer by checking if it makes sense.

Question 9. In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.
Answer:
Let a be the first term and r be the common ratio of given G.P.

Now, sum of first three terms = \( S_3 = \frac{a(r^3 - 1)}{r - 1} \)
Now, sum of first six terms = \( S_6 = \frac{a(r^6 - 1)}{r - 1} \)

It is given that
\( \frac{a(r^3 - 1)}{r - 1} \div \frac{a(r^6 - 1)}{r - 1} = \frac{125}{152} \)

\( \Rightarrow \frac{r^3 - 1}{r^6 - 1} = \frac{125}{152} \)

\( \Rightarrow \frac{r^3 - 1}{(r^3)^2 - (1)^2} = \frac{125}{152} \)

\( \Rightarrow \frac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \frac{125}{152} \)

\( \Rightarrow \frac{1}{r^3 + 1} = \frac{125}{152} \)

\( \Rightarrow r^3 + 1 = \frac{152}{125} \)

\( \Rightarrow r^3 = \frac{152}{125} - 1 = \frac{152 - 125}{125} = \frac{27}{125} \)

\( \Rightarrow r^3 = \left(\frac{3}{5}\right)^3 \)

\( \Rightarrow r = \frac{3}{5} \)

Hence, the common ratio is \( \frac{3}{5} \).
In simple words: We used the ratio of sums to create an equation. We simplified using the formula a² - b² = (a-b)(a+b) and solved for r.

📝 Teacher's Note: Explain that r⁶ - 1 = (r³)² - 1² which is a difference of squares. Students often miss this factorization step.

🎯 Exam Tip: Write the sum formulas first. Show the factorization step clearly. Check your answer by substituting back.

Question 10. Find how many terms of G.P. \( \frac{2}{9} - \frac{1}{3} + \frac{1}{2} \) ……. must be added to get the sum equal to \( \frac{55}{72} \)?
Answer:
Given G.P.: \( \frac{2}{9} - \frac{1}{3} + \frac{1}{2} \) ………
Here,
First term, \( a = \frac{2}{9} \)

Common ratio, \( r = \frac{-\frac{1}{3}}{\frac{2}{9}} = -\frac{3}{2} \) < 1

Let required number of terms be n.
\( \Rightarrow S_n = \frac{55}{72} \)

\( \Rightarrow \frac{a(1 - r^n)}{1 - r} = \frac{55}{72} \)

\( \Rightarrow \frac{\frac{2}{9}\left[1 - \left(\frac{-3}{2}\right)^n\right]}{1 - \left(\frac{-3}{2}\right)} = \frac{55}{72} \)

\( \Rightarrow \frac{\frac{2}{9}\left[1 - \left(\frac{-3}{2}\right)^n\right]}{\frac{5}{2}} = \frac{55}{72} \)

\( \Rightarrow \frac{2}{9}\left[1 - \left(\frac{-3}{2}\right)^n\right] = \frac{55}{72} \times \frac{5}{2} \)

\( \Rightarrow 1 - \left(\frac{-3}{2}\right)^n = \frac{55}{72} \times \frac{5}{2} \times \frac{9}{2} \)

\( \Rightarrow 1 - \left(\frac{-3}{2}\right)^n = \frac{275}{32} \)

\( \Rightarrow 1 - \frac{275}{32} = \left(\frac{-3}{2}\right)^n \)

\( \Rightarrow -\frac{243}{32} = \left(\frac{-3}{2}\right)^n \)

\( \Rightarrow \left(-\frac{3}{2}\right)^5 = \left(\frac{-3}{2}\right)^n \)

\( \Rightarrow n = 5 \)

∴ Required number of terms = 5
In simple words: This G.P. has negative terms alternating with positive terms. We used the sum formula and solved step by step to find n = 5.

📝 Teacher's Note: Show students that when r is negative, the terms alternate in sign. Check that (-3/2)⁵ = -243/32 by calculating step by step.

🎯 Exam Tip: For negative common ratio, be careful with signs. Use brackets around negative fractions. Show each calculation step clearly.

Question 11. If the sum 1 + 2 + 2² + ………. + 2ⁿ⁻¹ is 255, find the value of n.
Answer:
Required series: 1 + 2 + 2² + …… + 2ⁿ⁻¹
Now, \( \frac{2}{1} = 2, \frac{2^2}{2} = 2 \)
Thus, given series is a G.P. with
first term, a = 1
common ratio, r = 2 (r > 1)
Last term, l = 2ⁿ⁻¹
Let there be n terms in the series.
Then, \( S_n = 255 \)

\( \Rightarrow \frac{lr - a}{r - 1} = 255 \)
\( \Rightarrow \frac{2^{n-1} \times 2 - 1}{2 - 1} = 255 \)
\( \Rightarrow 2^{n-1} \times 2 - 1 = 255 \)
\( \Rightarrow 2^{n-1} \times 2 = 256 \)
\( \Rightarrow 2^{n-1} = 128 \)
\( \Rightarrow 2^{n-1} = 2^7 \)
\( \Rightarrow n - 1 = 7 \)
\( \Rightarrow n = 8 \)
In simple words: We recognized this as a G.P. with powers of 2. We used the sum formula and found that 256 = 2⁸, so n = 8.

📝 Teacher's Note: Students should know powers of 2 up to 2¹⁰ = 1024. Write them on the board: 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64, 2⁷ = 128, 2⁸ = 256.

🎯 Exam Tip: Memorize powers of 2 and 3 for quick solving. Always verify your answer by checking if it makes sense.

Question 12. Find the geometric mean between:
(i) \( \frac{4}{9} \) and \( \frac{9}{4} \)
(ii) 14 and \( \frac{7}{32} \)
(iii) 2a and 8a³

Solution 12(i).
Answer:
Geometric mean between \( \frac{4}{9} \) and \( \frac{9}{4} = \sqrt{\frac{4}{9} \times \frac{9}{4}} = \sqrt{1} = 1 \)
In simple words: Geometric mean is the square root of the product of two numbers. Here, the fractions multiply to give 1, so the mean is 1.

📝 Teacher's Note: Show students that 4/9 × 9/4 = (4×9)/(9×4) = 36/36 = 1. This makes the calculation very easy.

🎯 Exam Tip: Write the formula: Geometric mean = √(a×b). Always simplify the product before taking the square root.

Solution 12(ii).
Answer:
Geometric mean between 14 and \( \frac{7}{32} = \sqrt{14 \times \frac{7}{32}} = \sqrt{\frac{98}{32}} = \sqrt{\frac{49}{16}} = \frac{7}{4} = 1\frac{3}{4} \)
In simple words: We multiply the two numbers, simplify the fraction, then take the square root. 98÷2 = 49 and 32÷2 = 16, so we get 49/16.

📝 Teacher's Note: Show that √49 = 7 and √16 = 4. Also explain that 49/16 can be written as the mixed number 1¾.

🎯 Exam Tip: Simplify fractions before taking square roots. Check if the numerator and denominator are perfect squares.

Solution 12(iii).
Answer:
Geometric mean between 2a and 8a³ = \( \sqrt{2a \times 8a^3} = \sqrt{16a^4} = 4a^2 \)
In simple words: We multiply the terms: 2a × 8a³ = 16a⁴. Then we take the square root: √16 = 4 and √a⁴ = a².

📝 Teacher's Note: Remind students that when multiplying terms with same base, we add the powers: a¹ × a³ = a⁴. Also, √a⁴ = a² (taking half the power).

🎯 Exam Tip: For algebraic terms, multiply coefficients separately and add powers of variables. Remember √a²ⁿ = aⁿ.

Question 13. The sum of three numbers in G.P. is \( \frac{39}{10} \) and their product is 1. Find the numbers.
Answer:
Let the numbers be \( \frac{a}{r} \), a and ar.

\( \Rightarrow \frac{a}{r} \times a \times ar = 1 \)
\( \Rightarrow a^3 = 1 \)
\( \Rightarrow a = 1 \)

Now, \( \frac{a}{r} + a + ar = \frac{39}{10} \)
\( \Rightarrow \frac{1}{r} + 1 + r = \frac{39}{10} \)
\( \Rightarrow \frac{1 + r + r^2}{r} = \frac{39}{10} \)
\( \Rightarrow 10 + 10r + 10r^2 = 39r \)
\( \Rightarrow 10r^2 - 29r + 10 = 0 \)
\( \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \)
\( \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \)
\( \Rightarrow (2r - 5)(5r - 2) = 0 \)
\( \Rightarrow r = \frac{5}{2} \) or \( r = \frac{2}{5} \)

Thus, required terms are:
\( \frac{a}{r}, a, ar = \frac{1}{\frac{5}{2}}, 1, 1 \times \frac{5}{2} \) OR \( \frac{1}{\frac{2}{5}}, 1, 1 \times \frac{2}{5} \)
= \( \frac{2}{5}, 1, \frac{5}{2} \) OR \( \frac{5}{2}, 1, \frac{2}{5} \)
In simple words: We assumed three numbers in G.P. form. Their product gave us a = 1. Their sum gave us a quadratic equation which we solved to find r.

📝 Teacher's Note: Explain that three numbers in G.P. are written as a/r, a, ar. This ensures the middle term squared equals the product of outer terms.

🎯 Exam Tip: Always start with a/r, a, ar for three numbers in G.P. Use the product condition first to find a, then use sum condition to find r.

Question 14. The first term of a G.P. is -3 and the square of the second term is equal to its 4th term. Find its 7th term.
Answer:
Let the G.P. be: a, ar, ar², ar³, ar⁴, ar⁵, ar⁶, ...
Given: a = -3
Second term = ar = -3r
Fourth term = ar³ = -3r³

Given condition: (Second term)² = Fourth term
\( \Rightarrow (-3r)^2 = -3r^3 \)
\( \Rightarrow 9r^2 = -3r^3 \)
\( \Rightarrow 9r^2 + 3r^3 = 0 \)
\( \Rightarrow 3r^2(3 + r) = 0 \)

Since r ≠ 0 (otherwise it won't be a G.P.),
\( \Rightarrow 3 + r = 0 \)
\( \Rightarrow r = -3 \)

Seventh term = ar⁶ = (-3) × (-3)⁶ = (-3) × 729 = -2187
In simple words: We set up the condition that (second term)² equals fourth term. This gave us an equation to find the common ratio r = -3. Then we calculated the 7th term.

📝 Teacher's Note: Show students that (-3)⁶ = 729 by calculating: (-3)² = 9, (-3)⁴ = 81, (-3)⁶ = 729. Negative number to even power gives positive result.

🎯 Exam Tip: Write the general terms clearly. Set up the equation from the given condition. Remember that (-3)⁶ is positive, but a = -3 is negative.

Question 15. Find the 5th term of the G.P. \( \frac{5}{2} \), 1, .....
Answer:
Given:
First term (a) = \( \frac{5}{2} \)

Step 1: Find the common ratio (r).
And, common ratio (r) = \( \frac{1}{\frac{5}{2}} = \frac{2}{5} \)

Step 2: Use the formula for nth term.
Now, \( t_n = ar^{n-1} \)

Step 3: Find the 5th term.
\( \Rightarrow 5^{th} \text{ term} = t_5 = \frac{5}{2} \times \left(\frac{2}{5}\right)^{5-1} = \frac{5}{2} \times \left(\frac{2}{5}\right)^4 = \left(\frac{2}{5}\right)^3 = \frac{8}{125} \)

5th term = \( \frac{8}{125} \)
In simple words: In a G.P., each term is found by multiplying the first term with the common ratio raised to a power. Here we multiply \( \frac{5}{2} \) by \( \left(\frac{2}{5}\right)^4 \) to get the 5th term.

📝 Teacher's Note: Show students that common ratio = second term ÷ first term. Then use the formula \( t_n = ar^{n-1} \). Students often forget to subtract 1 from n.

🎯 Exam Tip: Write the formula clearly first. Show each step. Write the final answer with proper fraction form. Check your arithmetic carefully.

Question 16. The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Answer:
Given:
First term (a) = 125

Step 1: Find the common ratio (r).
And, common ratio (r) = \( \frac{25}{125} = \frac{1}{5} \)

Step 2: Find the 5th term.
Now, \( t_n = ar^{n-1} \)
\( \Rightarrow 5^{th} \text{ term} = t_5 = 125 \times \left(\frac{1}{5}\right)^{5-1} = 125 \times \left(\frac{1}{5}\right)^4 = 125 \times \frac{1}{625} = \frac{1}{5} \)

Step 3: Find the 6th term.
\( \Rightarrow 6^{th} \text{ term} = t_6 = 125 \times \left(\frac{1}{5}\right)^{6-1} = 125 \times \left(\frac{1}{5}\right)^5 = 125 \times \frac{1}{3125} = \frac{1}{25} \)

5th term = \( \frac{1}{5} \) and 6th term = \( \frac{1}{25} \)
In simple words: We found that each term is 5 times smaller than the previous term. So we keep dividing by 5 to get the next terms.

📝 Teacher's Note: Emphasize that r = 1/5 means the terms get smaller. Students can verify by checking: 125, 25, 5, 1, 1/5, 1/25. Each term is the previous term ÷ 5.

🎯 Exam Tip: Always find r first by dividing second term by first term. Then apply the formula. Show both answers clearly with proper fractions.

Question 17. Find the sum of the sequence -\( \frac{1}{3} \), 1, -3, 9, ............ upto 8 terms.
Answer:
Given:
Here, \( \frac{1}{-\frac{1}{3}} = \frac{-3}{1} = \frac{9}{-3} = -3 \)

Step 1: Identify the sequence properties.
Thus, the given sequence is a G.P with first term (a) = -\( \frac{1}{3} \) and common ratio (r) = -3 (r < 1).

Step 2: Apply the sum formula.
Number of terms to be added, n = 8
\( \therefore S_n = \frac{a(1-r^n)}{1-r} \)

Step 3: Calculate the sum.
\( \Rightarrow S_8 = \frac{-\frac{1}{3}(1-(-3)^8)}{1-3} = \frac{-1+3^8}{12} = \frac{1}{12}(3^8-1) \)

Sum of 8 terms = \( \frac{1}{12}(3^8-1) \)
In simple words: This is a G.P. where terms alternate between positive and negative. We use the sum formula for finite G.P. because we want only 8 terms.

📝 Teacher's Note: Show students the pattern: negative, positive, negative, positive. The common ratio is negative (-3), which causes this alternating pattern. Check that |r| > 1 so we use the finite sum formula.

🎯 Exam Tip: Write "r = -3" clearly. Use the correct formula for |r| > 1. Calculate \( 3^8 = 6561 \) if needed. Final answer: \( \frac{6560}{12} = \frac{1640}{3} \).

Question 18. The first term of a G.P. in 27. If the 8thterm be \( \frac{1}{81} \), what will be the sum of 10 terms?
Answer:
Given:
First term, a = 27
8th term = \( ar^7 = \frac{1}{81} \)
n = 10

Step 1: Find the common ratio.
Now, \( \frac{ar^7}{a} = \frac{\frac{1}{81}}{27} \)
\( \Rightarrow r^7 = \frac{1}{2187} \)
\( \Rightarrow r^7 = \left(\frac{1}{3}\right)^7 \)
\( \Rightarrow r = \frac{1}{3} \) (r < 1)

Step 2: Apply the sum formula.
\( \therefore S_n = \frac{a(1-r^n)}{1-r} \)

Step 3: Calculate the sum of 10 terms.
\( \Rightarrow S_{10} = \frac{27\left(1-\left(\frac{1}{3}\right)^{10}\right)}{1-\frac{1}{3}} = \frac{27\left(1-\frac{1}{3^{10}}\right)}{\frac{2}{3}} = \frac{81}{2}\left(1-3^{-10}\right) \)

Sum of 10 terms = \( \frac{81}{2}(1-3^{-10}) \)
In simple words: We first found r by using the given 8th term. Then we used the sum formula for finite G.P. because r < 1.

📝 Teacher's Note: Help students see that \( 27 = 3^3 \) and \( \frac{1}{81} = \frac{1}{3^4} \). So \( r^7 = \frac{3^4}{3^3 \times 3^7} = \frac{1}{3^7} \). This gives r = 1/3.

🎯 Exam Tip: First find r using \( ar^{n-1} = \text{given term} \). Write r clearly. Use the correct sum formula for r < 1. Show all steps neatly.

Question 19. Find a G.P. for which the sum of first two terms is -4 and the fifth term is 4 times the third term.
Answer:
Step 1: Set up the G.P. terms.
Let the five terms of the given G.P. be \( \frac{a}{r^2}, \frac{a}{r} \), a, ar, \( ar^2 \)

Step 2: Use the first condition.
Given, sum of first two terms = -4
\( \frac{a}{r^2} + \frac{a}{r} = -4 \)
\( \Rightarrow \frac{a+ar}{r^2} = -4 \)
\( \Rightarrow a + ar = -4r^2 \)
\( \Rightarrow a(1+r) = -4r^2 \)
\( \Rightarrow a = \frac{-4r^2}{1+r} \)

Step 3: Use the second condition.
And, 5th term = 4(3rd term)
\( \Rightarrow ar^2 = 4(a) \)
\( \Rightarrow r^2 = 4 \)
\( \Rightarrow r = \pm 2 \)

Step 4: Find 'a' for each value of r.
When r = +2, \( a = -\frac{4(2)^2}{1+2} = -\frac{16}{3} \)
When r = -2, \( a = -\frac{4(-2)^2}{1-2} = 16 \)

Step 5: Write the required G.P.
Thus, the required terms are \( \frac{a}{r^2}, \frac{a}{r} \), a, ar, \( ar^2 \)

For r = +2: \( -\frac{16}{3} \div 4, -\frac{16}{3} \div 2, -\frac{16}{3}, -\frac{16}{3} \times 2, -\frac{16}{3} \times 4 \) OR \( -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, -\frac{32}{3}, -\frac{64}{3} \)
For r = -2: 4, -8, 16, -32, 64

The required G.P. are: \( -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, -\frac{32}{3}, -\frac{64}{3} \) OR 4, -8, 16, -32, 64
In simple words: We used algebra to find the first term and common ratio from the two given conditions. There are two possible G.P.s that satisfy both conditions.

📝 Teacher's Note: Explain that we can write any G.P. as \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \) with 'a' as the middle term. This makes calculations easier. Check both answers by verifying the given conditions.

🎯 Exam Tip: Set up the G.P. symmetrically around the middle term. Solve the two conditions step by step. Always check which values of r work. Write both possible G.P.s as your final answer.

Additional Questions

Question 1. Find the sum of n terms of the series:
(i) 4 + 44 + 444 + ........
(ii) 0.8 + 0.88 + 0.888 + .............

Solution 1(i).
Answer:
Required sum = 4 + 44 + 444 + ........ up to n terms
= 4{1 + 11 + 111 + .......... up to n terms}
= \( \frac{4}{9} \)(9 + 99 + 999 + ......... up to n terms)
= \( \frac{4}{9} \)[(10 - 1) + (100 - 1) + (1000 - 1) + ......... up to n terms]
= \( \frac{4}{9} \)[{10 + 10² + 10³ + ........ up to n terms} - {1 + 1 + 1 + ......up to n terms}]
= \( \frac{4}{9} \)[\( \frac{10(10^n - 1)}{10 - 1} \) - n]
= \( \frac{4}{9} \)[\( \frac{10}{9} \)(10ⁿ - 1) - n]

Sum = \( \frac{4}{9} \)[\( \frac{10}{9} \)(10ⁿ - 1) - n]
In simple words: We broke each term like 444 = 4 × 111 = 4 × (100 + 10 + 1). Then we used the G.P. formula to find the sum.

📝 Teacher's Note: Show students the pattern: 4 = 4×1, 44 = 4×11, 444 = 4×111. Factor out 4 first. Then work with the series 1, 11, 111 which can be written as 9, 99, 999 divided by 9.

🎯 Exam Tip: Factor out the common part first (here it's 4). Convert to G.P. form using 9, 99, 999 = 9(1, 11, 111). Use G.P. sum formula correctly. Show all steps clearly.

Solution 1(ii).
Answer:
Required sum = 0.8 + 0.88 + 0.888 + ........ up to n terms
= 8{0.1 + 0.11 + 0.111 + .......... up to n terms}
= \( \frac{8}{9} \)(0.9 + 0.99 + 0.999 + .........up to n terms)
= \( \frac{8}{9} \)[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + ......... up to n terms]
= \( \frac{8}{9} \)[(1 + 1 + 1 + ........ up to n terms) - (0.1 + 0.01 + 0.001 + ...... up to n terms)]
= \( \frac{8}{9} \)[(1 + 1 + 1 + ........ up to n terms) - (\( \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \)..... up to n terms)]
= \( \frac{8}{9} \)[n - \( \frac{1}{10} \)(1 - \( \frac{1}{10^n} \))] [\( \therefore r = \frac{1}{10} < 1 \)]
= \( \frac{8}{9} \)[n - \( \frac{10}{9} \times \frac{1}{10} \)(1 - \( \frac{1}{10^n} \))]
= \( \frac{8}{9} \)[n - \( \frac{1}{9} \)(1 - \( \frac{1}{10^n} \))]

Sum = \( \frac{8}{9} \)[n - \( \frac{1}{9} \)(1 - \( \frac{1}{10^n} \))]
In simple words: We broke each decimal like 0.888 = 8 × 0.111. Then we used the pattern that 0.9 = 1 - 0.1, and so on.

📝 Teacher's Note: Help students see that 0.8 = 8×0.1, 0.88 = 8×0.11. Factor out 8. Then work with 0.1, 0.11, 0.111 series. Use the fact that 0.9 = 1 - 0.1, etc.

🎯 Exam Tip: Factor out 8 first. Convert 0.9, 0.99, 0.999 to (1-0.1), (1-0.01), (1-0.001). Use G.P. sum formula for the decimal part. Combine carefully.

Question 2. Find the sum of infinite terms of each of the following geometric progression:
(i) 1 + \( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} \) + ......
(ii) 1 - \( \frac{1}{2} + \frac{1}{4} - \frac{1}{8} \) + ......
(iii) \( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} \) + .....
(iv) \( \sqrt{2} - \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{1}{4\sqrt{2}} \) + ......
(v) \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \frac{1}{9\sqrt{3}} \) + ........
Answer:
(i) Here a = 1, r = \( \frac{1}{3} \) (|r| < 1)
Sum to infinity = \( \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \)

(ii) Here a = 1, r = -\( \frac{1}{2} \) (|r| < 1)
Sum to infinity = \( \frac{1}{1-(-\frac{1}{2})} = \frac{1}{1+\frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \)

(iii) Here a = \( \frac{1}{3} \), r = \( \frac{1}{3} \) (|r| < 1)
Sum to infinity = \( \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \)

(iv) Here a = \( \sqrt{2} \), r = \( \frac{-\frac{1}{\sqrt{2}}}{\sqrt{2}} = -\frac{1}{2} \) (|r| < 1)
Sum to infinity = \( \frac{\sqrt{2}}{1-(-\frac{1}{2})} = \frac{\sqrt{2}}{\frac{3}{2}} = \frac{2\sqrt{2}}{3} \)

(v) Here a = \( \sqrt{3} \), r = \( \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3} \) (|r| < 1)
Sum to infinity = \( \frac{\sqrt{3}}{1-\frac{1}{3}} = \frac{\sqrt{3}}{\frac{2}{3}} = \frac{3\sqrt{3}}{2} \)

In simple words: For infinite G.P. sum, we use the formula \( \frac{a}{1-r} \) only when |r| < 1. This means the terms get smaller and smaller, so we can find a finite sum.

📝 Teacher's Note: Emphasize that infinite sum exists only when |r| < 1. Show students how to find r by dividing second term by first term. Check that |r| < 1 before using the formula.

🎯 Exam Tip: Always write "a = ..." and "r = ..." clearly. Check |r| < 1. Use formula \( \frac{a}{1-r} \). Simplify fractions completely. Show all working steps for full marks.

Solution 2(i).
Answer:
Given G.P.: \( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \)
Here,
First term, \( a = 1 \)
Common ratio, \( r = \frac{1/3}{1} = \frac{1}{3} \) \( \left( |r| = \left|\frac{1}{3}\right| = \frac{1}{3} < 1 \right) \)
∴ Required sum = \( \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} \)
In simple words: This is an infinite G.P. where each term is one-third of the previous term. Since the common ratio is less than 1, the sum converges to 3/2.

📝 Teacher's Note: Show students how to identify the common ratio by dividing any term by the previous term. Make sure they check |r| < 1 before using the infinite sum formula.

🎯 Exam Tip: Always write |r| < 1 to show the series converges. Then use the formula a/(1-r). Show all steps clearly.

Solution 2(ii).
Answer:
Given G.P.: \( 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \ldots \)
Here,
First term, \( a = 1 \)
Common ratio, \( r = \frac{-1/2}{1} = -\frac{1}{2} \) \( \left( |r| = \left|-\frac{1}{2}\right| = \frac{1}{2} < 1 \right) \)
∴ Required sum = \( \frac{a}{1-r} = \frac{1}{1-\left(-\frac{1}{2}\right)} = \frac{1}{1+\frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \)
In simple words: This G.P. has alternating positive and negative terms. The common ratio is -1/2, so terms get smaller and change sign. The sum is 2/3.

📝 Teacher's Note: Emphasize that negative common ratio means alternating signs. Students often forget the negative sign when calculating 1-r.

🎯 Exam Tip: Write r = -1/2 clearly. Then write 1-r = 1-(-1/2) = 1+1/2 = 3/2. Show each step to avoid sign errors.

Solution 2(iii).
Answer:
Given G.P.: \( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \)
Here,
First term, \( a = \frac{1}{3} \)
Common ratio, \( r = \frac{1/3^2}{1/3} = \frac{1}{3} \) \( \left( |r| = \left|\frac{1}{3}\right| = \frac{1}{3} < 1 \right) \)
∴ Required sum = \( \frac{a}{1-r} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2} \)
In simple words: Each term is one-third of the previous term. Starting with 1/3, the infinite sum equals 1/2.

📝 Teacher's Note: Students often confuse this with the series starting from 1. Point out that the first term here is 1/3, not 1.

🎯 Exam Tip: Identify the first term carefully. Here a = 1/3, not 1. Use the formula correctly with the right first term.

Solution 2(iv).
Answer:
Given G.P.: \( \sqrt{2} - \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} - \frac{1}{4\sqrt{2}} + \ldots \)
Here,
First term, \( a = \sqrt{2} \)
Common ratio, \( r = \frac{-1/\sqrt{2}}{\sqrt{2}} = -\frac{1}{2} \) \( \left( |r| = \left|-\frac{1}{2}\right| = \frac{1}{2} < 1 \right) \)
∴ Required sum = \( \frac{a}{1-r} = \frac{\sqrt{2}}{1-\left(-\frac{1}{2}\right)} = \frac{\sqrt{2}}{1+\frac{1}{2}} = \frac{\sqrt{2}}{\frac{3}{2}} = \frac{2\sqrt{2}}{3} \)
In simple words: This G.P. has alternating signs and involves square roots. The sum converges to 2√2/3.

📝 Teacher's Note: Help students see that -1/√2 ÷ √2 = -1/2. Work through the division of surds step by step.

🎯 Exam Tip: Be careful with surd division. Write each step clearly. Don't forget to rationalize if needed.

Solution 2(v).
Answer:
Given G.P.: \( \sqrt{3} + \frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \frac{1}{9\sqrt{3}} + \ldots \)
Here,
First term, \( a = \sqrt{3} \)
Common ratio, \( r = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3} \) \( \left( |r| = \left|\frac{1}{3}\right| = \frac{1}{3} < 1 \right) \)
∴ Required sum = \( \frac{a}{1-r} = \frac{\sqrt{3}}{1-\frac{1}{3}} = \frac{\sqrt{3}}{\frac{2}{3}} = \frac{3\sqrt{3}}{2} \)
In simple words: Each term is one-third of the previous term. Starting with √3, the infinite sum is 3√3/2.

📝 Teacher's Note: Show students that 1/√3 ÷ √3 = 1/3. Practice surd division with several examples.

🎯 Exam Tip: Check your common ratio calculation carefully with surds. Write r = 1/3 clearly and verify |r| < 1.

Question 3. The second term of a G.P. is 9 and sum of its infinite terms is 48. Find its first three terms.
Answer:
Let a be the first term and r be the common ratio of a G.P.
2nd term, \( t_2 = ar = 9 \)
\( \Rightarrow r = \frac{9}{a} \)
Sum of its infinite terms, S = 48
\( \Rightarrow \frac{a}{1-r} = 48 \)
\( \Rightarrow \frac{a}{1-\frac{9}{a}} = 48 \)
\( \Rightarrow \frac{a^2}{a-9} = 48 \)
\( \Rightarrow a^2 = 48a - 432 \)
\( \Rightarrow a^2 - 48a + 432 = 0 \)
\( \Rightarrow a^2 - 36a - 12a + 432 = 0 \)
\( \Rightarrow a(a - 36) - 12(a - 36) = 0 \)
\( \Rightarrow (a - 36)(a - 12) = 0 \)
\( \Rightarrow a = 36 \text{ or } a = 12 \)

When \( a = 36, r = \frac{9}{36} = \frac{1}{4} \)
\( \Rightarrow \) 1st term = 36,
2nd term = \( ar = 36 \times \frac{1}{4} = 9 \)
3rd term = \( ar^2 = 36 \times \frac{1}{16} = \frac{9}{4} \)

When \( a = 12, r = \frac{9}{12} = \frac{3}{4} \)
\( \Rightarrow \) 1st term = 12,
2nd term = \( ar = 12 \times \frac{3}{4} = 9 \)
3rd term = \( ar^2 = 12 \times \frac{9}{16} = \frac{27}{4} \)

The first three terms are either: 36, 9, 9/4 or 12, 9, 27/4
In simple words: We use the given conditions to make equations. Solving gives us two possible sets of first three terms.

📝 Teacher's Note: Explain that both solutions are valid since we get a quadratic equation. Check both by substituting back into the original conditions.

🎯 Exam Tip: Always check |r| < 1 for infinite sum to exist. Both r = 1/4 and r = 3/4 satisfy this condition. Show both solutions clearly.

Question 4. Find three geometric means between 1/3 and 432.
Answer:
Let \( G_1, G_2, G_3 \) be three geometric means between \( a = \frac{1}{3} \) and \( b = 432 \).
Then, \( \frac{1}{3}, G_1, G_2, G_3, 432 \) is a G.P.
Thus, we have
First term = \( a = \frac{1}{3} \)
5th term of the G.P. = \( ar^4 = 432 \)
\( \Rightarrow \frac{1}{3} \times r^4 = 432 \)
\( \Rightarrow r^4 = 1296 \)
\( \Rightarrow r^4 = 6^4 \)
\( \Rightarrow r = 6 \)
\( \therefore G_1 = ar = \frac{1}{3} \times 6 = 2 \)
\( G_2 = ar^2 = \frac{1}{3} \times 6 \times 6 = 12 \)
\( G_3 = ar^3 = \frac{1}{3} \times 6 \times 6 \times 6 = 72 \)

The three geometric means are: 2, 12, 72
In simple words: We find the common ratio by using the formula for the 5th term. Then we calculate each geometric mean using ar, ar², ar³.

📝 Teacher's Note: Emphasize that n geometric means between two numbers create a G.P. with (n+2) terms total. Here 3 means create 5 terms.

🎯 Exam Tip: Write the complete G.P. sequence first. Then identify which term number the last given term is. Use ar^(n-1) formula correctly.

Question 5. Find:
(i) two geometric means between 2 and 16
(ii) four geometric means between 3 and 96
(iii) five geometric means between 32/9 and 81/2

Solution 5(i).
Answer:
Let \( G_1, G_2 \) be two geometric means between \( a = 2 \) and \( b = 16 \).
Then, \( 2, G_1, G_2, 16 \) is a G.P.
Thus, we have
First term = \( a = 2 \)
4th term of the G.P. = \( ar^3 = 16 \)
\( \Rightarrow 2 \times r^3 = 16 \)
\( \Rightarrow r^3 = 8 \)
\( \Rightarrow r^3 = 2^3 \)
\( \Rightarrow r = 2 \)
\( \therefore G_1 = ar = 2 \times 2 = 4 \)
\( G_2 = ar^2 = 2 \times 2 \times 2 = 8 \)

The two geometric means are: 4, 8
In simple words: Between 2 and 16, we insert two numbers that make a G.P. Each term is double the previous one.

📝 Teacher's Note: Point out that 2, 4, 8, 16 is a familiar sequence where each term doubles. This helps students see the pattern.

🎯 Exam Tip: Verify your answer by checking if it forms a G.P.: 2, 4, 8, 16. Each ratio should be the same (here, 2).

Solution 5(ii).
Answer:
Let \( G_1, G_2, G_3, G_4 \) be four geometric means between \( a = 3 \) and \( b = 96 \).
Then, \( 3, G_1, G_2, G_3, G_4, 96 \) is a G.P.
Thus, we have
First term = \( a = 3 \)
6th term of the G.P. = \( ar^5 = 96 \)
\( \Rightarrow 3 \times r^5 = 96 \)
\( \Rightarrow r^5 = 32 \)
\( \Rightarrow r^5 = 2^5 \)
\( \Rightarrow r = 2 \)
\( \therefore G_1 = ar = 3 \times 2 = 6 \)
\( G_2 = ar^2 = 3 \times 4 = 12 \)
\( G_3 = ar^3 = 3 \times 8 = 24 \)
\( G_4 = ar^4 = 3 \times 16 = 48 \)

The four geometric means are: 6, 12, 24, 48
In simple words: We insert four numbers between 3 and 96. Each number is double the previous one, creating a G.P.

📝 Teacher's Note: Show that the sequence 3, 6, 12, 24, 48, 96 has common ratio 2. Each term doubles as we move forward.

🎯 Exam Tip: Count carefully: 4 geometric means create a 6-term G.P. So the 6th term = ar⁵. Don't confuse the position with the number of means.

Solution 5(iii).
Answer:
Let \( G_1, G_2, G_3, G_4, G_5 \) be five geometric means between
\( a = 3\frac{5}{9} = \frac{32}{9} \) and \( b = 40\frac{1}{2} = \frac{81}{2} \).
Then, \( \frac{32}{9}, G_1, G_2, G_3, G_4, G_5, \frac{81}{2} \) is a G.P.
Thus, we have
First term = \( a = \frac{32}{9} \)
7th term of the G.P. = \( ar^6 = \frac{81}{2} \)
\( \Rightarrow \frac{32}{9} \times r^6 = \frac{81}{2} \)
\( \Rightarrow r^6 = \frac{81}{2} \times \frac{9}{32} = \frac{729}{64} \)
\( \Rightarrow r^6 = \left(\frac{3}{2}\right)^6 \)
\( \Rightarrow r = \frac{3}{2} \)

\( G_1 = ar = \frac{32}{9} \times \frac{3}{2} = \frac{16}{3} \)
\( G_2 = ar^2 = \frac{32}{9} \times \frac{9}{4} = 8 \)
\( G_3 = ar^3 = \frac{32}{9} \times \frac{27}{8} = 12 \)
\( G_4 = ar^4 = \frac{32}{9} \times \frac{81}{16} = 18 \)
\( G_5 = ar^5 = \frac{32}{9} \times \frac{243}{32} = 27 \)

The five geometric means are: 16/3, 8, 12, 18, 27
In simple words: We insert five numbers between two fractions. The common ratio is 3/2, so each term is 1.5 times the previous term.

📝 Teacher's Note: Help students convert mixed numbers to improper fractions first. Show fraction multiplication step by step to avoid errors.

🎯 Exam Tip: Convert mixed numbers correctly at the start. Check your final answer by verifying the common ratio is consistent throughout the sequence.

Question 6. The sum of three numbers in G.P. is \( \frac{39}{10} \) and their product is 1. Find the numbers.
Answer:
Given:
Sum of three numbers in G.P. = \( \frac{39}{10} \) and their product = 1

Step 1: Let the numbers be \( \frac{a}{r} \), a, ar

Step 2: Use the product condition.
\( \frac{a}{r} \times a \times ar = 1 \)
\( \implies a^3 = 1 \)
\( \implies a = 1 \)

Step 3: Use the sum condition.
\( \frac{a}{r} + a + ar = \frac{39}{10} \)
Since a = 1:
\( \frac{1}{r} + 1 + r = \frac{39}{10} \)

Step 4: Solve for r.
\( \frac{1}{r} + 1 + r = \frac{39}{10} \)
\( r + \frac{1}{r} = \frac{39}{10} - 1 = \frac{39-10}{10} = \frac{29}{10} \)
\( r^2 + 1 = \frac{29}{10}r \)
\( 10r^2 + 10 = 29r \)
\( 10r^2 - 29r + 10 = 0 \)

Step 5: Factorize the quadratic.
\( 10r^2 - 4r - 25r + 10 = 0 \)
\( 2r(5r - 2) - 5(5r - 2) = 0 \)
\( (5r - 2)(2r - 5) = 0 \)

Either \( 5r - 2 = 0 \), then \( r = \frac{2}{5} \)
or \( 2r - 5 = 0 \), then \( r = \frac{5}{2} \)

Step 6: Find the numbers.
Numbers are \( \frac{2}{5} \), 1, \( \frac{4}{25} \), or \( \frac{5}{2} \), 1, \( \frac{25}{4} \)

In simple words: We found three numbers that multiply to give 1 and add to give \( \frac{39}{10} \). Since they are in G.P., we used the pattern \( \frac{a}{r} \), a, ar.

📝 Teacher's Note: Tell students that in G.P. problems, we often use \( \frac{a}{r} \), a, ar as the three terms. This makes calculations easier. Show them how the product becomes \( a^3 \).

🎯 Exam Tip: Always write "Let the three numbers be \( \frac{a}{r} \), a, ar" first. Then use both given conditions to get two equations. Solve step by step.

Question 7. Find the numbers in G.P. whose sum is 52 and the sum of whose product in pairs is 624.
Answer:
Given:
Sum of three numbers = 52
Sum of product in pairs = 624

Step 1: Let the numbers be a, ar and \( ar^2 \).
\( a + ar + ar^2 = 52 \) ...(i)

Step 2: Find sum of products in pairs.
\( (a \times ar) + (ar \times ar^2) + (ar^2 \times a) = 624 \)
\( a^2r + a^2r^3 + a^2r^2 = 624 \)
\( ar(a + ar^2 + ar) = 624 \)
\( ar \times 52 = 624 \) ...[From (i)]
\( ar = 12 \)
\( a = \frac{12}{r} \)

Step 3: Substitute in equation (i).
\( \frac{12}{r} + \frac{12}{r} \times r + \frac{12}{r} \times r^2 = 52 \)
\( \frac{12}{r} + 12 + 12r = 52 \)
\( \frac{12 + 12r + 12r^2}{r} = 52 \)
\( 12 + 12r + 12r^2 = 52r \)
\( 12r^2 - 40r + 12 = 0 \)
\( 3r^2 - 10r + 3 = 0 \)

Step 4: Solve the quadratic.
\( 3r^2 - 9r - r + 3 = 0 \)
\( 3r(r - 3) - 1(r - 3) = 0 \)
\( (3r - 1)(r - 3) = 0 \)

Either \( r = \frac{1}{3} \) or \( r = 3 \)

Step 5: Find the numbers.
When \( r = \frac{1}{3} \): \( a = \frac{12}{\frac{1}{3}} = 36 \)
Numbers are: 36, 12, 4

When \( r = 3 \): \( a = \frac{12}{3} = 4 \)
Numbers are: 4, 12, 36

In simple words: We found three numbers in G.P. that add to 52. When we multiply them in pairs and add those products, we get 624.

📝 Teacher's Note: Explain that "product in pairs" means multiply first×second, second×third, first×third, then add all three results. Use simple examples like 2, 4, 8.

🎯 Exam Tip: Write clearly what "sum of products in pairs" means. Show all three products: ab, bc, ca. Then add them. This gets you marks.

Question 8. The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.
Answer:
Given:
Sum of three numbers = 21
Sum of their squares = 189

Step 1: Let the numbers be a, ar and \( ar^2 \).
\( (a)^2 + (ar)^2 + (ar^2)^2 = 189 \)
\( a^2 + a^2r^2 + a^2r^4 = 189 \)

Step 2: Use the sum condition.
\( a + ar + ar^2 = 21 \)
\( (a + ar + ar^2)^2 = 21^2 \)
\( a^2 + a^2r^2 + a^2r^4 + 2a^2r + 2a^2r^3 + 2a^2r^2 = 441 \)
\( 189 + 2ar(a + ar^2 + ar) = 441 \)
\( 2ar \times 21 = 441 - 189 \)
\( 42ar = 252 \)
\( ar = 6 \)

Step 3: Find r.
\( r = \frac{6}{a} \)

Step 4: Substitute back.
\( a + ar + ar^2 = 21 \)
\( a + a \times \frac{6}{a} + a \times \frac{36}{a^2} = 21 \)
\( a + 6 + \frac{36}{a} = 21 \)
\( a^2 + 6a + 36 = 21a \)
\( a^2 - 15a + 36 = 0 \)
\( a^2 - 12a - 3a + 36 = 0 \)
\( a(a - 12) - 3(a - 12) = 0 \)
\( (a - 12)(a - 3) = 0 \)

Either a = 12 or a = 3

Step 5: Find the complete sets.
When a = 12: \( r = \frac{6}{12} = \frac{1}{2} \)
Numbers are: 12, 6, 3

When a = 3: \( r = \frac{6}{3} = 2 \)
Numbers are: 3, 6, 12

In simple words: We found three numbers in G.P. that add to 21. When we square each number and add them, we get 189.

📝 Teacher's Note: Show students that squaring the sum gives us extra terms with 2ar. This is the key step. Use the identity \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \).

🎯 Exam Tip: Always expand \( (a + ar + ar^2)^2 \) carefully. Write each step clearly. The term 2ar appears three times, so factor it out as \( 2ar(a + ar + ar^2) \).