GSEB Class 11 Maths Solutions Chapter 9 Sequences and Series Exercise 9.3

Question 1. Find the 20th and nth terms of the G.P., \( \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots \)
Answer: Here, the first term \( a = \frac{5}{2} \).
The common ratio \( r = \frac{5}{4} \div \frac{5}{2} = \frac{5}{4} \times \frac{2}{5} = \frac{1}{2} \).
The general formula for the \( n \)th term of a G.P. is \( T_n = ar^{n-1} \).
For the 20th term, \( n = 20 \):
\( T_{20} = a r^{20-1} = \frac{5}{2} \left(\frac{1}{2}\right)^{19} = \frac{5}{2 \times 2^{19}} = \frac{5}{2^{20}} \)
For the \( n \)th term, \( T_n = a r^{n-1} = \frac{5}{2} \left(\frac{1}{2}\right)^{n-1} = \frac{5}{2 \times 2^{n-1}} = \frac{5}{2^n} \)
In simple words: We found the starting number and how much it grows each step. Then, we used a general rule to find the 20th number and a rule for any number in the sequence.

Exam Tip: Remember to correctly identify the first term (a) and the common ratio (r) before applying the geometric progression formulas.

Question 2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
We are given that the 8th term \( T_8 = 192 \).
Using the formula for the \( n \)th term of a G.P., \( T_n = ar^{n-1} \):
\( T_8 = ar^{8-1} = ar^7 = 192 \dots (1) \)
We are also given that the common ratio \( r = 2 \).
Substitute \( r = 2 \) into equation (1):
\( a(2)^7 = 192 \)
\( a \times 128 = 192 \)
\( a = \frac{192}{128} = \frac{3 \times 64}{2 \times 64} = \frac{3}{2} \)
Now, we need to find the 12th term, \( T_{12} \).
\( T_{12} = ar^{12-1} = ar^{11} \)
Substitute the values of \( a = \frac{3}{2} \) and \( r = 2 \):
\( T_{12} = \frac{3}{2} (2)^{11} = \frac{3}{2} \times 2048 = 3 \times 1024 = 3072 \)
Alternatively, we could write \( T_{12} = ar^{11} = (ar^7)(r^4) \)
\( T_{12} = 192 \times 2^4 \quad [\because r=2] \)
\( T_{12} = 192 \times 16 = 3072 \)
In simple words: We first found the starting number using the given 8th term and common ratio. Then, we used these values to calculate the 12th number in the sequence.

Exam Tip: When finding a higher term given a lower term and the common ratio, you can use the relationship \( T_m = T_n \times r^{m-n} \) to simplify calculations, as shown in the alternative method.

Question 3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that \( q^2 = ps \).
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
According to the problem:
The 5th term \( T_5 = p \implies ar^4 = p \dots (1) \)
The 8th term \( T_8 = q \implies ar^7 = q \dots (2) \)
The 11th term \( T_{11} = s \implies ar^{10} = s \dots (3) \)
Now, we need to prove that \( q^2 = ps \).
Let's consider \( q^2 \):
\( q^2 = (ar^7)^2 \) [Using (2)]
\( q^2 = a^2 r^{14} \)
We can rewrite \( a^2 r^{14} \) as \( (ar^4)(ar^{10}) \)
From equations (1) and (3), we know that \( ar^4 = p \) and \( ar^{10} = s \).
Therefore, \( q^2 = ps \)
This proves the required relationship.
In simple words: We wrote out the 5th, 8th, and 11th numbers using the general G.P. formula. Then, we showed that the square of the 8th term is equal to the product of the 5th and 11th terms.

Exam Tip: This problem demonstrates a property of geometric progressions: if three terms are in G.P. and evenly spaced, the middle term is the geometric mean of the other two. Always express terms using \( ar^{n-1} \) and simplify to reveal such relationships.

Question 4. The 4th term of a G.P. is square of its 2nd term and the first term is – 3. Determine its 7th term?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
We are given the first term \( a = -3 \).
We are also given that the 4th term of the G.P. is the square of its 2nd term.
\( T_4 = (T_2)^2 \)
Using the formula \( T_n = ar^{n-1} \):
\( ar^{4-1} = (ar^{2-1})^2 \)
\( ar^3 = (ar)^2 \)
\( ar^3 = a^2 r^2 \)
Since \( a \neq 0 \) (as \( a = -3 \)), we can divide both sides by \( ar^2 \).
\( \frac{ar^3}{ar^2} = \frac{a^2 r^2}{ar^2} \)
\( r = a \)
Since \( a = -3 \), it implies that \( r = -3 \).
Now, we need to determine the 7th term, \( T_7 \).
\( T_7 = ar^{7-1} = ar^6 \)
Substitute the values \( a = -3 \) and \( r = -3 \):
\( T_7 = (-3)(-3)^6 \)
\( T_7 = (-3)^7 \)
\( T_7 = -2187 \)
In simple words: We used the given rules about the 4th and 2nd terms, plus the first term, to find the common growth factor. Once we knew that, we easily calculated the 7th number in the sequence.

Exam Tip: Be careful with signs when dealing with negative terms or common ratios. Remember that an odd power of a negative number is negative, and an even power is positive.

Question 5. Which term of the sequence:
(a) \( 2, 2\sqrt{2}, 4, \dots \) is 128?
(b) \( \sqrt{3}, 3, 3\sqrt{3}, \dots \) is 729?
(c) \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots \) is \( \frac{1}{19683} \)?
Answer:
(a) For the G.P. \( 2, 2\sqrt{2}, 4, \dots \):
The first term \( a = 2 \).
The common ratio \( r = \frac{2\sqrt{2}}{2} = \sqrt{2} \).
Let the \( n \)th term be 128.
\( T_n = ar^{n-1} = 128 \)
\( 2(\sqrt{2})^{n-1} = 128 \)
\( (\sqrt{2})^{n-1} = \frac{128}{2} \)
\( (\sqrt{2})^{n-1} = 64 \)
We know that \( \sqrt{2} = 2^{1/2} \) and \( 64 = 2^6 \).
\( (2^{1/2})^{n-1} = 2^6 \)
\( 2^{\frac{n-1}{2}} = 2^6 \)
Equating the exponents:
\( \frac{n-1}{2} = 6 \)
\( n-1 = 12 \)
\( n = 13 \)
Thus, the 13th term is 128.
(b) For the G.P. \( \sqrt{3}, 3, 3\sqrt{3}, \dots \):
Here, the first term \( a = \sqrt{3} \).
The common ratio \( r = \frac{3}{\sqrt{3}} = \sqrt{3} \).
Let the \( n \)th term be 729.
\( T_n = ar^{n-1} = 729 \)
\( (\sqrt{3})(\sqrt{3})^{n-1} = 729 \)
\( (\sqrt{3})^n = 729 \)
We know that \( 729 = 9^3 = (3^2)^3 = 3^6 \). Also \( \sqrt{3} = 3^{1/2} \).
\( (3^{1/2})^n = 3^6 \)
\( 3^{n/2} = 3^6 \)
Equating the exponents:
\( \frac{n}{2} = 6 \)
\( n = 12 \)
Hence, the 12th term is 729.
(c) For the G.P. \( \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots \):
Here, the first term \( a = \frac{1}{3} \).
The common ratio \( r = \frac{1/9}{1/3} = \frac{1}{9} \times \frac{3}{1} = \frac{1}{3} \).
Let the \( n \)th term be \( \frac{1}{19683} \).
\( T_n = ar^{n-1} = \frac{1}{19683} \)
\( \frac{1}{3} \left(\frac{1}{3}\right)^{n-1} = \frac{1}{19683} \)
\( \left(\frac{1}{3}\right)^n = \frac{1}{19683} \)
We know that \( 19683 = 3^9 \).
\( \left(\frac{1}{3}\right)^n = \frac{1}{3^9} \)
\( \left(\frac{1}{3}\right)^n = \left(\frac{1}{3}\right)^9 \)
Equating the exponents:
\( n = 9 \)
Therefore, the 9th term is \( \frac{1}{19683} \).
In simple words: For each sequence, we first found the starting number and the growth factor. Then, we used the formula for the nth term to figure out which position the given target number held in the sequence.

Exam Tip: When solving for 'n', make sure to express both sides of the equation with the same base before equating the exponents. This often involves powers of 2, 3, or other small prime numbers.

Question 6. For what values of x, the numbers \( \frac{-2}{7}, x, \frac{- 7}{2} \) are in G.P.?
Answer: For three numbers to be in a Geometric Progression (G.P.), the ratio of consecutive terms must be constant (the common ratio, c.r.).
So, if \( \frac{-2}{7}, x, \frac{-7}{2} \) are in G.P., then:
\( \frac{x}{\frac{-2}{7}} = \frac{\frac{-7}{2}}{x} \)
This is because each ratio equals the common ratio.
Cross-multiplying gives:
\( x^2 = \left(\frac{-2}{7}\right) \times \left(\frac{-7}{2}\right) \)
\( x^2 = 1 \)
Taking the square root of both sides:
\( x = \pm 1 \)
Thus, the possible values for \( x \) are 1 and -1.
In simple words: For numbers to be in a geometric sequence, the middle number squared must equal the product of the first and last numbers. We used this rule to find the value of x.

Exam Tip: Remember the property of a G.P. that if a, b, c are in G.P., then \( b^2 = ac \). This shortcut can save time in problems involving three consecutive terms.

Find the sum to indicated number of terms of each of the following geometric progressions in questions 7 to 10:

Question 7. 0.15, 0.015, 0.0015, \( \dots \) 20 terms.
Answer: For this geometric progression:
The first term \( a = 0.15 \).
The common ratio \( r = \frac{0.015}{0.15} = 0.1 \).
The number of terms \( n = 20 \).
Since \( |r| = |0.1| < 1 \), we use the sum formula for \( S_n \) when \( r < 1 \):
\( S_n = \frac{a(1 - r^n)}{1 - r} \)
Substitute the values:
\( S_{20} = \frac{0.15(1 - (0.1)^{20})}{1 - 0.1} \)
\( S_{20} = \frac{0.15(1 - (0.1)^{20})}{0.9} \)
\( S_{20} = \frac{15}{90} (1 - (0.1)^{20}) \)
\( S_{20} = \frac{1}{6} (1 - (0.1)^{20}) \)
In simple words: We found the starting number and how much it changes each step. Then, using a special rule for sums in a geometric sequence, we calculated the total for 20 terms.

Exam Tip: Always check if the common ratio \( r \) is less than 1 or greater than 1, as this determines which form of the sum formula \( S_n \) to use. Both forms are mathematically equivalent but one might be easier to calculate depending on \( r \).

Question 8. \( \sqrt{7}, \sqrt{21}, 3\sqrt{7}, \dots \) n terms.
Answer: For this geometric progression:
The first term \( a = \sqrt{7} \).
The common ratio \( r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{\frac{21}{7}} = \sqrt{3} \).
The number of terms is \( n \).
Since \( |r| = |\sqrt{3}| > 1 \), we use the sum formula for \( S_n \) when \( r > 1 \):
\( S_n = \frac{a(r^n - 1)}{r - 1} \)
Substitute the values:
\( S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)}{\sqrt{3} - 1} \)
To rationalize the denominator, multiply the numerator and denominator by \( (\sqrt{3} + 1) \):
\( S_n = \frac{\sqrt{7}((\sqrt{3})^n - 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( S_n = \frac{\sqrt{7}(\sqrt{3} + 1)((\sqrt{3})^n - 1)}{3 - 1} \)
\( S_n = \frac{\sqrt{7}(\sqrt{3} + 1)((\sqrt{3})^n - 1)}{2} \)
In simple words: We found the starting number and the multiplication factor for each step. Then, we used the sum rule for geometric sequences to write down the sum for 'n' terms, making sure to simplify the bottom part of the fraction.

Exam Tip: Always remember to rationalize the denominator in your final answer if it contains radicals. This is a standard practice in simplifying expressions.

Question 9. \( 1, -a, a^2, -a^3, \dots \) n terms (a \( \neq \) - 1).
Answer: For this geometric progression:
The first term \( a_{GP} = 1 \). (Using \( a_{GP} \) to distinguish from the variable \( a \)).
The common ratio \( r = \frac{-a}{1} = -a \).
The number of terms is \( n \).
We use the sum formula for \( S_n \):
\( S_n = \frac{a_{GP}(1 - r^n)}{1 - r} \)
Substitute the values \( a_{GP} = 1 \) and \( r = -a \):
\( S_n = \frac{1(1 - (-a)^n)}{1 - (-a)} \)
\( S_n = \frac{1 - (-a)^n}{1 + a} \)
In simple words: We identified the first number and the way each number is multiplied to get the next one. Then, we put these into the sum formula for 'n' terms.

Exam Tip: Be careful with signs when the common ratio is negative, especially when raising it to a power \( n \). \( (-a)^n \) will alternate in sign depending on whether \( n \) is even or odd.

Question 10. \( x^3, x^5, x^7, \dots \) n terms (if x \( \neq \pm \) 1).
Answer: For this geometric progression:
The first term \( a = x^3 \).
The common ratio \( r = \frac{x^5}{x^3} = x^2 \).
The number of terms is \( n \).
We use the sum formula for \( S_n \):
\( S_n = \frac{a(1 - r^n)}{1 - r} \)
Substitute the values \( a = x^3 \) and \( r = x^2 \):
\( S_n = \frac{x^3(1 - (x^2)^n)}{1 - x^2} \)
\( S_n = \frac{x^3(1 - x^{2n})}{1 - x^2} \)
This formula is valid if \( x^2 \neq 1 \), which means \( x \neq \pm 1 \), as stated in the question.
In simple words: We found the starting term and the common multiplier from the sequence. Then, we applied the standard sum formula for 'n' terms in a geometric progression.

Exam Tip: Always make sure to simplify exponents correctly, for example, \( (x^m)^n = x^{mn} \). Also, pay attention to any restrictions on variables, like \( x \neq \pm 1 \) in this case, as they ensure the denominator is not zero.

Question 11. Evaluate \( \sum_{k=1}^{11} (2 + 3k) \).
Answer: The sum \( \sum_{k=1}^{11} (2 + 3k) \) means we need to sum the terms \( (2 + 3k) \) for \( k \) from 1 to 11.
Let's list the first few terms:
When \( k=1 \): \( 2 + 3(1) = 5 \)
When \( k=2 \): \( 2 + 3(2) = 8 \)
When \( k=3 \): \( 2 + 3(3) = 11 \)
The sequence is \( 5, 8, 11, \dots \)
This is an Arithmetic Progression (A.P.) with:
First term \( a = 5 \)
Common difference \( d = 8 - 5 = 3 \)
Number of terms \( n = 11 \)
The sum of an A.P. is given by \( S_n = \frac{n}{2} [2a + (n-1)d] \).
\( S_{11} = \frac{11}{2} [2(5) + (11-1)3] \)
\( S_{11} = \frac{11}{2} [10 + (10)3] \)
\( S_{11} = \frac{11}{2} [10 + 30] \)
\( S_{11} = \frac{11}{2} [40] \)
\( S_{11} = 11 \times 20 \)
\( S_{11} = 220 \)
Alternatively, we can separate the sum:
\( \sum_{k=1}^{11} (2 + 3k) = \sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3k \)
\( = (2 \times 11) + 3 \sum_{k=1}^{11} k \)
The sum of the first \( n \) natural numbers is \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \).
So, \( \sum_{k=1}^{11} k = \frac{11(11+1)}{2} = \frac{11 \times 12}{2} = 11 \times 6 = 66 \).
\( = 22 + 3(66) \)
\( = 22 + 198 \)
\( = 220 \)
The original OCR also shows a different approach, possibly misinterpreting the sum as a geometric progression or a more complex series of operations.
The expression \( (2 + 3)1 + (2 + 3)2 + (2 + 33) + \dots + (2 + 311) \) seems to be a mistranscription or an incorrect interpretation of the sum.
Let's follow the OCR's provided final answer if it matches a valid interpretation of the initial series.
The OCR text shows \( 22 + \frac{3(3^{11}-1)}{3-1} \). This form implies a sum of a geometric progression, which is not what \( 2+3k \) generates.
Given \( \sum_{k=1}^{11} (2 + 3k) \):
This means we are adding terms like \( (2+3), (2+6), (2+9), \dots, (2+33) \).
This is an arithmetic progression \( 5, 8, 11, \dots, 35 \).
The sum \( S_{11} = \frac{11}{2} (5+35) = \frac{11}{2}(40) = 220 \).
The OCR's derivation `22 + \frac{3}{2}(3^{11} – 1)` would result from a sum like `\sum_{k=1}^{11} (2 + 3^k)`, which is different from the question.
Therefore, I will provide the correct arithmetic progression solution.
In simple words: We added up the numbers in the sequence. First, we found the pattern, which was an arithmetic sequence (adding the same number each time). Then, we used the formula for the sum of an arithmetic sequence to get the final total.

Exam Tip: Carefully identify whether a sequence is arithmetic (constant difference) or geometric (constant ratio) before applying sum formulas. Misidentifying the sequence type is a common error.

Question 12. The sum of first three terms of a G.P. is \( \frac{39}{10} \) and their product is 1. Find the common ratio and the terms?
Answer: Let the first three terms of the G.P. be \( \frac{a}{r}, a, ar \).
Given that the product of these three terms is 1:
\( \left(\frac{a}{r}\right) \times a \times (ar) = 1 \)
\( a^3 = 1 \)
This implies \( a = 1 \).
Given that the sum of these three terms is \( \frac{39}{10} \):
\( \frac{a}{r} + a + ar = \frac{39}{10} \)
Substitute \( a = 1 \):
\( \frac{1}{r} + 1 + r = \frac{39}{10} \)
To remove the fractions, multiply the entire equation by \( 10r \):
\( 10 + 10r + 10r^2 = 39r \)
Rearrange the terms into a standard quadratic equation:
\( 10r^2 + 10r - 39r + 10 = 0 \)
\( 10r^2 - 29r + 10 = 0 \)
Factorize the quadratic equation. We need two numbers that multiply to \( 10 \times 10 = 100 \) and add to -29. These numbers are -25 and -4.
\( 10r^2 - 25r - 4r + 10 = 0 \)
\( 5r(2r - 5) - 2(2r - 5) = 0 \)
\( (2r - 5)(5r - 2) = 0 \)
This gives two possible values for the common ratio \( r \):
\( 2r - 5 = 0 \implies r = \frac{5}{2} \)
\( 5r - 2 = 0 \implies r = \frac{2}{5} \)
Case 1: If \( r = \frac{5}{2} \) and \( a = 1 \).
The terms of the G.P. are \( \frac{a}{r}, a, ar \)
\( \frac{1}{5/2}, 1, 1 \times \frac{5}{2} \)
\( \frac{2}{5}, 1, \frac{5}{2} \)
Case 2: If \( r = \frac{2}{5} \) and \( a = 1 \).
The terms of the G.P. are \( \frac{a}{r}, a, ar \)
\( \frac{1}{2/5}, 1, 1 \times \frac{2}{5} \)
\( \frac{5}{2}, 1, \frac{2}{5} \)
Both sets of terms are valid and are just in reverse order of each other.
In simple words: We used the product of the terms to find the middle term, 'a'. Then, we used the sum of the terms to set up an equation to find the common growth factor, 'r'. Finally, we listed the three terms for both possible values of 'r'.

Exam Tip: When dealing with an odd number of terms in a G.P., assuming the terms as \( \frac{a}{r}, a, ar \) (or \( \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2 \) for 5 terms) often simplifies the calculation for the product, allowing you to directly find 'a'.

Question 13. How many terms of G.P. \( 3, 3^2, 3^3, \dots \) are needed to give the sum 120?
Answer: Let \( n \) be the number of terms of the G.P. \( 3, 3^2, 3^3, \dots \) whose sum is 120.
From the given G.P.:
The first term \( a = 3 \).
The common ratio \( r = \frac{3^2}{3} = 3 \).
The sum of \( n \) terms \( S_n = 120 \).
Since \( |r| = 3 > 1 \), we use the sum formula:
\( S_n = \frac{a(r^n - 1)}{r - 1} \)
Substitute the values:
\( 120 = \frac{3(3^n - 1)}{3 - 1} \)
\( 120 = \frac{3(3^n - 1)}{2} \)
Multiply both sides by \( \frac{2}{3} \):
\( 120 \times \frac{2}{3} = 3^n - 1 \)
\( 40 \times 2 = 3^n - 1 \)
\( 80 = 3^n - 1 \)
\( 80 + 1 = 3^n \)
\( 81 = 3^n \)
We know that \( 81 = 3^4 \).
So, \( 3^4 = 3^n \)
Equating the exponents:
\( n = 4 \)
Therefore, 4 terms of the given G.P. are needed to give the sum 120.
In simple words: We found the starting number and how much it grows each time. Then, we used the sum formula for a geometric sequence to figure out how many terms we needed to add to reach a total of 120.

Exam Tip: When solving for 'n' in geometric progression problems, you'll often end up with an exponential equation. Express both sides with the same base to easily equate the exponents and solve for 'n'.

Question 14. The sum of the first three terms of G.P. is 16 and the sum of the next three terms is 128. Determine the first term, common ratio and the sum to n terms of the G.P?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio.
Given that the sum of the first three terms is 16:
\( S_3 = \frac{a(1 - r^3)}{1 - r} = 16 \dots (1) \)
Given that the sum of the next three terms (i.e., terms 4, 5, 6) is 128.
The sum of the first six terms \( S_6 = S_3 + (\text{sum of terms 4, 5, 6}) = 16 + 128 = 144 \).
So, \( S_6 = \frac{a(1 - r^6)}{1 - r} = 144 \dots (2) \)
Divide equation (2) by equation (1):
\( \frac{\frac{a(1 - r^6)}{1 - r}}{\frac{a(1 - r^3)}{1 - r}} = \frac{144}{16} \)
\( \frac{1 - r^6}{1 - r^3} = 9 \)
Use the algebraic identity \( 1 - r^6 = (1 - r^3)(1 + r^3) \):
\( \frac{(1 - r^3)(1 + r^3)}{1 - r^3} = 9 \)
\( 1 + r^3 = 9 \)
\( r^3 = 9 - 1 \)
\( r^3 = 8 \)
\( r^3 = 2^3 \)
This means the common ratio \( r = 2 \).
Now, substitute \( r = 2 \) back into equation (1) to find \( a \):
\( \frac{a(1 - 2^3)}{1 - 2} = 16 \)
\( \frac{a(1 - 8)}{-1} = 16 \)
\( \frac{a(-7)}{-1} = 16 \)
\( 7a = 16 \)
\( a = \frac{16}{7} \)
Finally, find the sum to \( n \) terms of the G.P.
Since \( r = 2 > 1 \), use the formula \( S_n = \frac{a(r^n - 1)}{r - 1} \).
\( S_n = \frac{\frac{16}{7}(2^n - 1)}{2 - 1} \)
\( S_n = \frac{16}{7}(2^n - 1) \)
In simple words: We used the given sums of terms to create two equations. By dividing these equations, we found the common growth factor. Then, we used this factor to calculate the first term. Finally, we wrote the general formula for the sum of 'n' terms.

Exam Tip: When given sums of consecutive blocks of terms in a G.P., dividing the sums often helps simplify the problem by eliminating 'a' and \( (1-r) \) and allowing you to solve for 'r' more easily.

Question 15. Given a G.P. with a = 729 and 7th term 64, determine S7.
Answer: We are given:
The first term \( a = 729 \).
The 7th term \( T_7 = 64 \).
Let \( r \) be the common ratio. Using the formula \( T_n = ar^{n-1} \):
\( T_7 = ar^{7-1} = ar^6 = 64 \)
Substitute \( a = 729 \):
\( 729 r^6 = 64 \)
\( r^6 = \frac{64}{729} \)
We know that \( 64 = 2^6 \) and \( 729 = 3^6 \).
\( r^6 = \frac{2^6}{3^6} \)
\( r^6 = \left(\frac{2}{3}\right)^6 \)
Therefore, \( r = \frac{2}{3} \).
Since \( |r| = \frac{2}{3} < 1 \), we use the sum formula for \( S_n \):
\( S_n = \frac{a(1 - r^n)}{1 - r} \)
We need to determine \( S_7 \). Substitute \( n = 7 \), \( a = 729 \), and \( r = \frac{2}{3} \):
\( S_7 = \frac{729 \left(1 - \left(\frac{2}{3}\right)^7\right)}{1 - \frac{2}{3}} \)
\( S_7 = \frac{729 \left(1 - \frac{2^7}{3^7}\right)}{\frac{1}{3}} \)
\( S_7 = 3 \times 729 \left(1 - \frac{128}{2187}\right) \)
\( S_7 = 2187 \left(\frac{2187 - 128}{2187}\right) \)
\( S_7 = 2187 \left(\frac{2059}{2187}\right) \)
\( S_7 = 2059 \)
In simple words: We used the given first term and the 7th term to find the common growth factor. Once we knew that, we used the sum formula for a geometric sequence to calculate the total of the first seven terms.

Exam Tip: Be careful when simplifying fractions involving powers; \( \frac{2^6}{3^6} = (\frac{2}{3})^6 \). Also, remember to choose the correct sum formula based on whether \( |r| < 1 \) or \( |r| > 1 \).

Question 16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
Given that the sum of the first two terms is -4:
\( S_2 = a + ar = a(1 + r) = -4 \dots (1) \)
Given that the fifth term is 4 times the third term:
\( T_5 = 4 T_3 \)
Using the formula \( T_n = ar^{n-1} \):
\( ar^4 = 4(ar^2) \)
Since \( a \neq 0 \) (otherwise the terms would all be zero) and \( r \neq 0 \) (otherwise the terms would be a, 0, 0, ... which is not a proper G.P. if sum is -4), we can divide both sides by \( ar^2 \):
\( r^2 = 4 \)
Taking the square root:
\( r = \pm 2 \)
We have two possible values for \( r \).
Case 1: \( r = 2 \).
Substitute \( r = 2 \) into equation (1):
\( a(1 + 2) = -4 \)
\( 3a = -4 \)
\( a = -\frac{4}{3} \)
The G.P. is \( a, ar, ar^2, \dots \):
\( -\frac{4}{3}, -\frac{4}{3}(2), -\frac{4}{3}(2^2), \dots \)
\( -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots \)
Case 2: \( r = -2 \).
Substitute \( r = -2 \) into equation (1):
\( a(1 + (-2)) = -4 \)
\( a(1 - 2) = -4 \)
\( a(-1) = -4 \)
\( a = 4 \)
The G.P. is \( a, ar, ar^2, \dots \):
\( 4, 4(-2), 4(-2)^2, 4(-2)^3, \dots \)
\( 4, -8, 16, -32, 64, \dots \)
Both are valid geometric progressions satisfying the given conditions.
In simple words: We used the relationship between the 5th and 3rd terms to find the common growth factor. This gave us two possible values for 'r'. Then, we used the sum of the first two terms to find the starting number 'a' for each possible 'r', listing both geometric sequences.

Exam Tip: Always consider both positive and negative roots when solving for \( r^2 = k \), as both might yield valid geometric progressions. Test each case to find the corresponding 'a' and the complete G.P.

Question 17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
According to the problem:
The 4th term \( T_4 = x \implies ar^{4-1} = ar^3 = x \dots (1) \)
The 10th term \( T_{10} = y \implies ar^{10-1} = ar^9 = y \dots (2) \)
The 16th term \( T_{16} = z \implies ar^{16-1} = ar^{15} = z \dots (3) \)
For \( x, y, z \) to be in G.P., the ratio of consecutive terms must be equal, i.e., \( \frac{y}{x} = \frac{z}{y} \). This also means \( y^2 = xz \).
Let's check the ratio \( \frac{y}{x} \):
\( \frac{y}{x} = \frac{ar^9}{ar^3} = r^{9-3} = r^6 \)
Now, let's check the ratio \( \frac{z}{y} \):
\( \frac{z}{y} = \frac{ar^{15}}{ar^9} = r^{15-9} = r^6 \)
Since \( \frac{y}{x} = r^6 \) and \( \frac{z}{y} = r^6 \), it means \( \frac{y}{x} = \frac{z}{y} \).
Therefore, \( x, y, z \) are in G.P.
In simple words: We wrote out the 4th, 10th, and 16th terms using the general formula for a geometric sequence. Then, we showed that the ratio between consecutive terms (y/x and z/y) was the same, which proves they form a geometric sequence.

Exam Tip: This problem illustrates a key property: if terms \( T_p, T_q, T_r \) of a G.P. are themselves in G.P., then the exponents \( p, q, r \) must be in A.P., i.e., \( q-p = r-q \). In this case, \( 10-4 = 6 \) and \( 16-10 = 6 \), confirming the result.

Question 18. Find the sum of first n terms of the sequence 8, 88, 888, 8888, \( \dots \).
Answer: Let \( S \) be the sum of the first \( n \) terms of the sequence \( 8, 88, 888, 8888, \dots \).
\( S = 8 + 88 + 888 + 8888 + \dots + \text{to n terms} \)
Factor out 8:
\( S = 8(1 + 11 + 111 + 1111 + \dots + \text{to n terms}) \)
Multiply and divide by 9:
\( S = \frac{8}{9}(9 + 99 + 999 + 9999 + \dots + \text{to n terms}) \)
Rewrite each term inside the parenthesis as a power of 10 minus 1:
\( S = \frac{8}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + (10^4 - 1) + \dots + \text{to n terms}] \)
Group the powers of 10 and the -1s:
\( S = \frac{8}{9}[(10 + 10^2 + 10^3 + 10^4 + \dots + \text{to n terms}) - (1 + 1 + 1 + 1 + \dots + \text{n times})] \)
The first part \( (10 + 10^2 + \dots + 10^n) \) is a geometric progression with \( a = 10 \), \( r = 10 \), and \( n \) terms. The sum is \( \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9} \).
The second part is \( (1 + 1 + \dots + \text{n times}) = n \).
Substitute these back into the expression for \( S \):
\( S = \frac{8}{9}\left[\frac{10(10^n - 1)}{9} - n\right] \)
In simple words: We changed each number in the sequence into a form that involved powers of 10. Then, we used the sum rule for a geometric sequence to add up the powers of 10 and subtracted the 'n' ones to get the final formula for the sum.

Exam Tip: Sequences like 8, 88, 888... are common. The trick is to factor out the common digit (e.g., 8), then multiply and divide by 9 to convert the terms into \( (10^k - 1) \) form, which can then be summed as a geometric progression minus a simple arithmetic sum.

Question 19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, \( \frac{1}{2} \).
Answer: Let the first sequence be \( A = (a_1, a_2, a_3, a_4, a_5) = (2, 4, 8, 16, 32) \).
Let the second sequence be \( B = (b_1, b_2, b_3, b_4, b_5) = (128, 32, 8, 2, \frac{1}{2}) \).
We need to find the sum of the products of the corresponding terms, which is \( S = a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 + a_5b_5 \).
Calculate each product:
\( a_1b_1 = 2 \times 128 = 256 \)
\( a_2b_2 = 4 \times 32 = 128 \)
\( a_3b_3 = 8 \times 8 = 64 \)
\( a_4b_4 = 16 \times 2 = 32 \)
\( a_5b_5 = 32 \times \frac{1}{2} = 16 \)
Now, sum these products:
\( S = 256 + 128 + 64 + 32 + 16 \)
This new sequence of products is \( 256, 128, 64, 32, 16 \). This is a Geometric Progression.
For this G.P.:
The first term \( a' = 256 \).
The common ratio \( r' = \frac{128}{256} = \frac{1}{2} \).
The number of terms \( n = 5 \).
Since \( |r'| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a'(1 - (r')^n)}{1 - r'} \):
\( S_5 = \frac{256 \left(1 - \left(\frac{1}{2}\right)^5\right)}{1 - \frac{1}{2}} \)
\( S_5 = \frac{256 \left(1 - \frac{1}{32}\right)}{\frac{1}{2}} \)
\( S_5 = 256 \times 2 \times \left(\frac{32 - 1}{32}\right) \)
\( S_5 = 512 \times \frac{31}{32} \)
\( S_5 = 16 \times 31 \)
\( S_5 = 496 \)
In simple words: We multiplied each number from the first list by the matching number from the second list. This created a new list of numbers that also followed a geometric pattern. Then, we added up all the numbers in this new list using the sum formula.

Exam Tip: When asked to find the sum of products of corresponding terms of two G.P.s, always check if the new sequence formed by the products is also a G.P. If it is, you can use the G.P. sum formula, which is often faster than adding terms individually.

Question 20. Show that the products of the corresponding terms of the sequences a, ar, ar², ... \( ar^{n-1} \) and A, AR, AR², ... \( AR^{n-1} \) form a G.P. and find the common ratio?
Answer: Let the first geometric sequence be \( G_1 = (a, ar, ar^2, \dots, ar^{n-1}) \).
Its first term is \( a \) and common ratio is \( r \).
Let the second geometric sequence be \( G_2 = (A, AR, AR^2, \dots, AR^{n-1}) \).
Its first term is \( A \) and common ratio is \( R \).
The products of the corresponding terms are:
\( T_1' = a \times A = aA \)
\( T_2' = ar \times AR = aARr \)
\( T_3' = ar^2 \times AR^2 = aA(rR)^2 \)
\( \dots \)
\( T_n' = ar^{n-1} \times AR^{n-1} = aA(rR)^{n-1} \)
The new sequence formed by these products is \( aA, aA(rR), aA(rR)^2, \dots, aA(rR)^{n-1} \).
To show that this sequence is a G.P., we need to demonstrate that the ratio of consecutive terms is constant.
Consider the ratio of the second term to the first term:
\( \frac{T_2'}{T_1'} = \frac{aARr}{aA} = rR \)
Consider the ratio of the third term to the second term:
\( \frac{T_3'}{T_2'} = \frac{aA(rR)^2}{aA(rR)} = rR \)
Since the ratio of consecutive terms is constant, \( rR \), the new sequence is a Geometric Progression.
The first term of this new G.P. is \( aA \).
The common ratio of this new G.P. is \( rR \).
In simple words: When you multiply terms that are in two different geometric sequences, the new sequence you get is also a geometric sequence. The new first term is just the product of the original first terms, and the new growth factor is the product of the original growth factors.

Exam Tip: This is a useful property to remember: the product of two geometric progressions is also a geometric progression. The first term is the product of their first terms, and the common ratio is the product of their common ratios.

Question 21. Find the numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the fourth term by 18?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
The terms are \( T_1 = a, T_2 = ar, T_3 = ar^2, T_4 = ar^3 \).
Given that the third term is greater than the first term by 9:
\( T_3 = T_1 + 9 \)
\( ar^2 = a + 9 \dots (1) \)
Given that the second term is greater than the fourth term by 18:
\( T_2 = T_4 + 18 \)
\( ar = ar^3 + 18 \dots (2) \)
From equation (2), rearrange the terms:
\( ar - ar^3 = 18 \)
\( ar(1 - r^2) = 18 \)
From equation (1), rearrange to solve for \( a \):
\( ar^2 - a = 9 \)
\( a(r^2 - 1) = 9 \)
\( -a(1 - r^2) = 9 \)
\( a(1 - r^2) = -9 \dots (3) \)
Now, divide equation (2) by equation (3):
\( \frac{ar(1 - r^2)}{a(1 - r^2)} = \frac{18}{-9} \)
\( r = -2 \)
Now substitute \( r = -2 \) back into equation (1):
\( a(-2)^2 = a + 9 \)
\( 4a = a + 9 \)
\( 3a = 9 \)
\( a = 3 \)
Now we have \( a = 3 \) and \( r = -2 \). The terms of the G.P. are:
\( T_1 = a = 3 \)
\( T_2 = ar = 3(-2) = -6 \)
\( T_3 = ar^2 = 3(-2)^2 = 3(4) = 12 \)
\( T_4 = ar^3 = 3(-2)^3 = 3(-8) = -24 \)
The required terms of the geometric progression are \( 3, -6, 12, -24 \).
In simple words: We used the given relationships between different terms to set up two equations involving the first term 'a' and the common growth factor 'r'. Solving these equations helped us find 'a' and 'r', which then let us list the numbers in the sequence.

Exam Tip: Systematically writing down equations based on the given conditions is crucial. Look for ways to simplify or combine equations, such as dividing them, to eliminate variables and solve for 'a' and 'r'.

Question 22. If the pth, qth and rth terms of a G.P. are a, b, c respectively, prove that \( a^{q-r}b^{r-p}c^{p-q} = 1 \).
Answer: Let \( A \) be the initial term and \( R \) be the common ratio of the G.P.
Given that:
The pth term is \( a \implies AR^{p-1} = a \dots (1) \)
The qth term is \( b \implies AR^{q-1} = b \dots (2) \)
The rth term is \( c \implies AR^{r-1} = c \dots (3) \)
We need to prove \( a^{q-r}b^{r-p}c^{p-q} = 1 \).
Substitute the expressions for \( a, b, c \) from (1), (2), (3) into the expression:
L.H.S. \( = (AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q} \)
Apply the power rule \( (x^m)^n = x^{mn} \):
\( = A^{q-r} (R^{p-1})^{q-r} \times A^{r-p} (R^{q-1})^{r-p} \times A^{p-q} (R^{r-1})^{p-q} \)
\( = A^{q-r} R^{(p-1)(q-r)} \times A^{r-p} R^{(q-1)(r-p)} \times A^{p-q} R^{(r-1)(p-q)} \)
Group the terms with base \( A \) and terms with base \( R \):
\( = A^{(q-r) + (r-p) + (p-q)} \times R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)} \)
Let's evaluate the exponent of \( A \):
\( (q-r) + (r-p) + (p-q) = q - r + r - p + p - q = 0 \)
So, the term with \( A \) becomes \( A^0 \).
Now, let's evaluate the exponent of \( R \):
\( (p-1)(q-r) = pq - pr - q + r \)
\( (q-1)(r-p) = qr - qp - r + p \)
\( (r-1)(p-q) = rp - rq - p + q \)
Summing these three parts:
\( (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) \)
\( = pq - pr - q + r + qr - pq - r + p + pr - qr - p + q \)
All terms cancel out:
\( = (pq - pq) + (-pr + pr) + (-q + q) + (r - r) + (qr - qr) + (p - p) = 0 \)
So, the term with \( R \) becomes \( R^0 \).
Therefore, the expression simplifies to:
\( A^0 \times R^0 = 1 \times 1 = 1 \)
L.H.S. = R.H.S., hence the proof is complete.
In simple words: We substituted the general forms of the pth, qth, and rth terms into the expression. By carefully combining the powers of the first term and the common ratio, all the exponents canceled out, showing the whole expression equals one.

Exam Tip: When dealing with such proofs involving powers of terms, the strategy is usually to express all terms using the formula \( AR^{n-1} \), then group the base \( A \) terms and base \( R \) terms, and finally show that their exponents sum to zero. Careful algebraic expansion and cancellation are key.

Question 23. If the first and the nth terms of a G.P. are a and b respectively and if p is the product of first n terms, prove that \( p^2 = (ab)^n \).
Answer: Let \( a_1 \) be the initial term and \( r \) be the common ratio of the G.P.
Given that the first term is \( a \), so \( a_1 = a \).
Given that the \( n \)th term is \( b \), so \( T_n = a_1 r^{n-1} = ar^{n-1} = b \dots (1) \)
Let \( p \) be the product of the first \( n \) terms of the G.P.:
\( p = T_1 \times T_2 \times T_3 \times \dots \times T_n \)
\( p = a \times (ar) \times (ar^2) \times \dots \times (ar^{n-1}) \)
Group all the \( a \) terms and all the \( r \) terms:
There are \( n \) terms of \( a \), so \( a \times a \times \dots \times a = a^n \).
The powers of \( r \) are \( 0, 1, 2, \dots, (n-1) \).
So, \( p = a^n \times r^{0+1+2+\dots+(n-1)} \)
The sum of the first \( (n-1) \) natural numbers is \( \frac{(n-1)n}{2} \).
So, \( p = a^n r^{\frac{n(n-1)}{2}} \)
Now, we need to prove \( p^2 = (ab)^n \). Let's square \( p \):
\( p^2 = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 \)
\( p^2 = a^{2n} r^{2 \times \frac{n(n-1)}{2}} \)
\( p^2 = a^{2n} r^{n(n-1)} \)
We can rewrite this as \( p^2 = (a^2 r^{n-1})^n \). This isn't quite right.
Let's try to express it in terms of \( a \) and \( b \).
From equation (1), \( b = ar^{n-1} \).
Consider \( ab \):
\( ab = a \times (ar^{n-1}) = a^2 r^{n-1} \)
Now, let's look at \( (ab)^n \):
\( (ab)^n = (a^2 r^{n-1})^n \)
\( (ab)^n = (a^2)^n (r^{n-1})^n \)
\( (ab)^n = a^{2n} r^{n(n-1)} \)
Comparing this with our expression for \( p^2 \), we see that:
\( p^2 = a^{2n} r^{n(n-1)} = (ab)^n \)
Hence, \( p^2 = (ab)^n \) is proved.
In simple words: We wrote the product of the first 'n' terms using the first term and common ratio. Then, we squared this product. Separately, we multiplied the first and 'n'th terms together and raised that product to the power of 'n'. We found that both results were the same, proving the relationship.

Exam Tip: This is a classic G.P. proof. The key insight is to express the product of terms in terms of \( a \) and \( r \), and then manipulate it to match the \( (ab)^n \) form, using the definition of the \( n \)th term \( b \).

Question 24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th terms is \( \frac{1}{r^n} \)?
Answer: Let \( a \) be the initial term and \( r \) be the common ratio of the G.P.
The sum of the first \( n \) terms of a G.P. is:
\( S_n = \frac{a(1 - r^n)}{1 - r} \dots (1) \)
The sum of terms from the \( (n+1) \)th term to the \( (2n) \)th term means we are summing \( n \) terms, starting from \( T_{n+1} \).
The \( (n+1) \)th term is \( T_{n+1} = ar^{(n+1)-1} = ar^n \).
This new sequence of terms (from \( T_{n+1} \) to \( T_{2n} \)) is also a G.P. Its first term is \( ar^n \), its common ratio is \( r \), and it has \( n \) terms.
The sum of these \( n \) terms is \( S'_{n} = \frac{\text{first term of this subsequence}(1 - r^n)}{1 - r} \)
\( S'_{n} = \frac{ar^n(1 - r^n)}{1 - r} \dots (2) \)
Now, we need to find the ratio of \( S_n \) to \( S'_{n} \):
\( \frac{S_n}{S'_{n}} = \frac{\frac{a(1 - r^n)}{1 - r}}{\frac{ar^n(1 - r^n)}{1 - r}} \)
Cancel out the common factors \( a \), \( (1 - r^n) \), and \( (1 - r) \):
\( \frac{S_n}{S'_{n}} = \frac{1}{r^n} \)
Hence, the ratio is \( \frac{1}{r^n} \), as required.
In simple words: We wrote the formula for the sum of the first 'n' terms. Then, we wrote the formula for the sum of the *next* 'n' terms, treating the (n+1)th term as the new starting point. When we divided the first sum by the second sum, all parts canceled out except for \( \frac{1}{r^n} \).

Exam Tip: When dealing with sums of blocks of terms in a G.P., remember that a subsequence starting from \( T_k \) is also a G.P. with a first term \( T_k \) and the same common ratio. This simplifies the sum calculation for the later block of terms.

Question 25. If a, b, c are in G.P., show that \( (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2 \).
Answer: Let \( r \) be the common ratio of the G.P. for \( a, b, c, d \).
Then, we can express the terms in relation to \( a \):
\( b = ar \)
\( c = ar^2 \)
\( d = ar^3 \)
Now, let's evaluate the Left Hand Side (L.H.S.) of the equation:
L.H.S. \( = (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) \)
Substitute the expressions for \( b, c, d \):
\( L.H.S. = (a^2 + (ar)^2 + (ar^2)^2)((ar)^2 + (ar^2)^2 + (ar^3)^2) \)
\( L.H.S. = (a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6) \)
Factor out \( a^2 \) from the first parenthesis and \( a^2r^2 \) from the second parenthesis:
\( L.H.S. = a^2(1 + r^2 + r^4) \times a^2r^2(1 + r^2 + r^4) \)
\( L.H.S. = a^4r^2(1 + r^2 + r^4)^2 \)
Now, let's evaluate the Right Hand Side (R.H.S.) of the equation:
R.H.S. \( = (ab + bc + cd)^2 \)
Substitute the expressions for \( b, c, d \):
\( R.H.S. = (a(ar) + (ar)(ar^2) + (ar^2)(ar^3))^2 \)
\( R.H.S. = (a^2r + a^2r^3 + a^2r^5)^2 \)
Factor out \( a^2r \) from inside the parenthesis:
\( R.H.S. = (a^2r(1 + r^2 + r^4))^2 \)
Apply the power of a product rule:
\( R.H.S. = (a^2r)^2 (1 + r^2 + r^4)^2 \)
\( R.H.S. = a^4r^2 (1 + r^2 + r^4)^2 \)
Since L.H.S. = \( a^4r^2(1 + r^2 + r^4)^2 \) and R.H.S. = \( a^4r^2(1 + r^2 + r^4)^2 \), we have shown that L.H.S. = R.H.S.
Hence, the equality is proved.
In simple words: We expressed all terms (b, c, d) using the first term 'a' and the common growth factor 'r'. Then, we calculated both sides of the equation separately, simplifying them until they were identical, which proved the statement.

Exam Tip: For proofs involving terms in G.P., the most effective strategy is to express all terms in terms of the first term 'a' and the common ratio 'r'. This allows you to manipulate and simplify expressions algebraically to show equality.

Question 26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Answer: Let the two numbers to be inserted between 3 and 81 be \( G_1 \) and \( G_2 \).
The resulting sequence will be \( 3, G_1, G_2, 81 \). This sequence forms a G.P.
Let \( a \) be the first term and \( r \) be the common ratio of this G.P.
Here, the first term \( a = 3 \).
The fourth term \( T_4 = 81 \).
Using the formula \( T_n = ar^{n-1} \):
\( T_4 = ar^{4-1} = ar^3 = 81 \)
Substitute \( a = 3 \):
\( 3r^3 = 81 \)
\( r^3 = \frac{81}{3} \)
\( r^3 = 27 \)
We know that \( 27 = 3^3 \).
\( r^3 = 3^3 \)
Therefore, the common ratio \( r = 3 \).
Now we can find \( G_1 \) and \( G_2 \):
\( G_1 = T_2 = ar = 3 \times 3 = 9 \)
\( G_2 = T_3 = ar^2 = 3 \times 3^2 = 3 \times 9 = 27 \)
Thus, the two numbers to be inserted are 9 and 27.
The resulting G.P. is \( 3, 9, 27, 81 \).
In simple words: We needed to find two numbers that fit into a geometric sequence between 3 and 81. We used the first and last numbers to figure out the common growth factor. Once we knew the growth factor, we easily found the two missing numbers.

Exam Tip: When inserting 'n' geometric means between two numbers, the total number of terms in the sequence becomes \( n+2 \). Use this to correctly set up the equation for the common ratio.

Question 27. Find the value of n so that \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} \) may be that geometric mean between a and b?
Answer: The geometric mean (G.M.) between two numbers \( a \) and \( b \) is \( \sqrt{ab} \).
We are given that \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} \) is equal to the geometric mean \( \sqrt{ab} \).
So, \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab} \)
Rewrite \( \sqrt{ab} \) as \( a^{1/2} b^{1/2} \):
\( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = a^{1/2} b^{1/2} \)
Cross-multiply:
\( a^{n+1}+b^{n+1} = a^{1/2} b^{1/2} (a^n+b^n) \)
\( a^{n+1}+b^{n+1} = a^{1/2} b^{1/2} a^n + a^{1/2} b^{1/2} b^n \)
Apply exponent rules \( x^m x^k = x^{m+k} \):
\( a^{n+1}+b^{n+1} = a^{n + 1/2} b^{1/2} + a^{1/2} b^{n + 1/2} \)
Rearrange terms to group common bases:
\( a^{n+1} - a^{n + 1/2} b^{1/2} = a^{1/2} b^{n + 1/2} - b^{n+1} \)
Factor out common terms on each side:
On the left side, factor out \( a^{n+1/2} \):
\( a^{n+1/2} (a^{(n+1) - (n+1/2)} - b^{1/2}) = a^{n+1/2} (a^{1/2} - b^{1/2}) \)
On the right side, factor out \( b^{n+1/2} \):
\( b^{n+1/2} (a^{1/2} - b^{(n+1) - (n+1/2)}) = b^{n+1/2} (a^{1/2} - b^{1/2}) \)
So, the equation becomes:
\( a^{n+1/2} (a^{1/2} - b^{1/2}) = b^{n+1/2} (a^{1/2} - b^{1/2}) \)
Since \( a \) and \( b \) are different (otherwise G.M. is just a, which would mean \( a^{n+1}+a^{n+1} = a^{n+1}+a^{n+1} \) which is always true and doesn't define n uniquely), \( (a^{1/2} - b^{1/2}) \neq 0 \).
Therefore, we can divide both sides by \( (a^{1/2} - b^{1/2}) \):
\( a^{n+1/2} = b^{n+1/2} \)
Divide both sides by \( b^{n+1/2} \):
\( \frac{a^{n+1/2}}{b^{n+1/2}} = 1 \)
\( \left(\frac{a}{b}\right)^{n+1/2} = 1 \)
For this equation to hold, the exponent must be zero, as \( \frac{a}{b} \neq 1 \) (since \( a \neq b \)).
\( n + \frac{1}{2} = 0 \)
\( n = -\frac{1}{2} \)
In simple words: We set the given expression equal to the formula for the geometric mean. Then, we did some algebra to rearrange and simplify the equation. By canceling common factors, we found that the exponent of \( \frac{a}{b} \) must be zero, which allowed us to solve for 'n'.

Exam Tip: This type of problem often involves equating expressions to a mean (arithmetic or geometric). The key is to carefully manipulate exponents and factor common terms. Remember that if \( x^y = 1 \) and \( x \neq 1 \), then \( y \) must be 0.

Question 28. The sum of two numbers is 6 times their geometric mean. Show that the numbers are in the ratio \( 3 + 2\sqrt{2}: 3 – 2\sqrt{2} \).
Answer: Let the two numbers be \( a \) and \( b \).
Their sum is \( a + b \).
Their geometric mean (G.M.) is \( \sqrt{ab} \).
According to the problem, the sum of two numbers is 6 times their geometric mean:
\( a + b = 6\sqrt{ab} \)
Divide both sides by \( 2\sqrt{ab} \):
\( \frac{a + b}{2\sqrt{ab}} = \frac{6\sqrt{ab}}{2\sqrt{ab}} \)
\( \frac{a + b}{2\sqrt{ab}} = \frac{3}{1} \)
Now, apply the componendo and dividendo rule. This rule states that if \( \frac{P}{Q} = \frac{R}{S} \), then \( \frac{P+Q}{P-Q} = \frac{R+S}{R-S} \).
Let \( P = a+b \) and \( Q = 2\sqrt{ab} \). Let \( R = 3 \) and \( S = 1 \).
\( \frac{(a+b) + 2\sqrt{ab}}{(a+b) - 2\sqrt{ab}} = \frac{3 + 1}{3 - 1} \)
Recognize that the numerator is \( (\sqrt{a} + \sqrt{b})^2 \) and the denominator is \( (\sqrt{a} - \sqrt{b})^2 \).
\( \frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{4}{2} \)
\( \left(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}\right)^2 = 2 \)
Take the square root of both sides:
\( \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \sqrt{2} \)
We can write \( \sqrt{2} \) as \( \frac{\sqrt{2}}{1} \). Apply componendo and dividendo again:
\( \frac{(\sqrt{a} + \sqrt{b}) + (\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b}) - (\sqrt{a} - \sqrt{b})} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \)
\( \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \)
\( \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \)
Square both sides to get the ratio of \( a \) to \( b \):
\( \frac{a}{b} = \left(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right)^2 \)
\( \frac{a}{b} = \frac{(\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2}{(\sqrt{2})^2 - 2(\sqrt{2})(1) + 1^2} \)
\( \frac{a}{b} = \frac{2 + 2\sqrt{2} + 1}{2 - 2\sqrt{2} + 1} \)
\( \frac{a}{b} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} \)
Hence, the numbers are in the ratio \( 3 + 2\sqrt{2}: 3 – 2\sqrt{2} \).
In simple words: We used the given relationship between the sum and geometric mean of two numbers. By applying a special algebraic rule called componendo and dividendo twice, we were able to simplify the expressions and find the ratio of the numbers as requested.

Exam Tip: Componendo and dividendo is a powerful tool for problems involving ratios, especially when sums and differences are involved. Recognize the perfect square forms \( (\sqrt{a} \pm \sqrt{b})^2 \) in the numerator and denominator after the first application.

Question 29. If A and G be A.M. and G.M. respectively between two positive numbers, prove that the numbers are \( A \pm \sqrt{(A + G)(A – G)} \).
Answer: Let the two positive numbers be \( a \) and \( b \).
The Arithmetic Mean (A.M.) is given by \( A = \frac{a+b}{2} \dots (1) \)
The Geometric Mean (G.M.) is given by \( G = \sqrt{ab} \). Squaring both sides, \( G^2 = ab \dots (2) \)
From equation (1), \( a+b = 2A \).
We know that \( (a-b)^2 = (a+b)^2 - 4ab \).
Substitute \( (a+b) = 2A \) and \( ab = G^2 \) into this identity:
\( (a-b)^2 = (2A)^2 - 4G^2 \)
\( (a-b)^2 = 4A^2 - 4G^2 \)
\( (a-b)^2 = 4(A^2 - G^2) \)
Taking the square root of both sides:
\( a-b = \pm \sqrt{4(A^2 - G^2)} \)
\( a-b = \pm 2\sqrt{A^2 - G^2} \dots (3) \)
We have two linear equations involving \( a \) and \( b \):
(i) \( a+b = 2A \)
(ii) \( a-b = \pm 2\sqrt{A^2 - G^2} \)
To find \( a \), add (i) and (ii):
\( (a+b) + (a-b) = 2A \pm 2\sqrt{A^2 - G^2} \)
\( 2a = 2A \pm 2\sqrt{A^2 - G^2} \)
Divide by 2:
\( a = A \pm \sqrt{A^2 - G^2} \)
To find \( b \), subtract (ii) from (i):
\( (a+b) - (a-b) = 2A \mp 2\sqrt{A^2 - G^2} \)
\( 2b = 2A \mp 2\sqrt{A^2 - G^2} \)
Divide by 2:
\( b = A \mp \sqrt{A^2 - G^2} \)
So the two numbers are \( A + \sqrt{A^2 - G^2} \) and \( A - \sqrt{A^2 - G^2} \).
We can also rewrite \( \sqrt{A^2 - G^2} \) as \( \sqrt{(A-G)(A+G)} \).
Therefore, the numbers are \( A \pm \sqrt{(A-G)(A+G)} \).
In simple words: We used the definitions of Arithmetic Mean and Geometric Mean to create equations for the sum and product of the two numbers. By solving these equations simultaneously, we could express the original numbers in terms of their mean values, which matched the desired form.

Exam Tip: This proof involves solving a system of equations where \( a+b \) and \( ab \) are known. The identity \( (a-b)^2 = (a+b)^2 - 4ab \) is very useful here. Remember that \( A^2 - G^2 \) can be factored as \( (A-G)(A+G) \).

Question 30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Answer: This situation represents a Geometric Progression (G.P.) because the number of bacteria doubles (multiplies by 2) every hour.
The original number of bacteria is the initial term, \( a = 30 \).
Since the bacteria double every hour, the common ratio is \( r = 2 \).
The number of bacteria after \( n \) hours can be found using the formula \( T_{n+1} = ar^n \), where \( n \) is the number of hours passed. (Or \( T_n = ar^{n-1} \) where \( n \) represents time, so \( T_1 \) is initial, \( T_2 \) after 1 hr, etc.). Let's use \( N_h \) for number after h hours, so \( N_h = a \cdot r^h \).
Number of bacteria at the start (0th hour) = \( 30 \).
Number of bacteria after 1st hour = \( 30 \times 2 = 60 \).
Number of bacteria after 2nd hour:
\( N_2 = a r^2 = 30 \times 2^2 = 30 \times 4 = 120 \)
Number of bacteria after 4th hour:
\( N_4 = a r^4 = 30 \times 2^4 = 30 \times 16 = 480 \)
Number of bacteria after \( n \)th hour:
\( N_n = a r^n = 30 \times 2^n \)
In simple words: We recognized that the bacteria growth followed a geometric pattern where the number doubled each hour. We used the starting count and the doubling rate to calculate the number of bacteria after 2 hours, 4 hours, and then for any number of 'n' hours.

Exam Tip: Problems involving exponential growth (like doubling, tripling, etc.) can often be modeled as geometric progressions. The initial amount is 'a' and the growth factor (e.g., 2 for doubling) is 'r'. Ensure you use the correct exponent for the number of time periods passed.

Question 31. What will? Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest of 10% compounded annually?
Answer: This is a problem of compound interest. The formula for the amount \( A \) after \( T \) years with principal \( P \), and annual interest rate \( r \) (as a decimal) compounded annually is:
\( A = P(1 + r)^T \)
In this problem, we are given:
Principal amount \( P = \text{Rs. } 500 \).
Annual interest rate \( r = 10\% = 0.10 \).
Time period \( T = 10 \) years.
Substitute these values into the compound interest formula:
\( A = 500(1 + 0.10)^{10} \)
\( A = 500(1.1)^{10} \)
To calculate \( (1.1)^{10} \):
\( (1.1)^1 = 1.1 \)
\( (1.1)^2 = 1.21 \)
\( (1.1)^3 = 1.331 \)
\( (1.1)^4 = 1.4641 \)
\( (1.1)^5 = 1.61051 \)
\( (1.1)^{10} = ((1.1)^5)^2 = (1.61051)^2 \approx 2.59374246 \)
So, \( A = 500 \times 2.59374246 \)
\( A = 1296.87123 \)
Therefore, the amount will be approximately Rs. 1296.87 after 10 years.
In simple words: We used the compound interest formula to find the final amount. We put in the starting money, the interest rate, and how many years it would grow. Then, we calculated the final value, which represents the initial deposit plus all the earned interest over the period.

Exam Tip: Remember the compound interest formula \( A = P(1 + r/n)^{nt} \). For annually compounded interest, \( n=1 \). Be careful with decimal conversions for percentage rates and perform exponentiation accurately, typically using a calculator for higher powers.

Question 32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation?
Answer: Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation.
Given that the Arithmetic Mean (A.M.) of the roots is 8:
\( \frac{\alpha + \beta}{2} = 8 \)
\( \alpha + \beta = 2 \times 8 = 16 \)
Given that the Geometric Mean (G.M.) of the roots is 5:
\( \sqrt{\alpha \beta} = 5 \)
Square both sides:
\( \alpha \beta = 5^2 = 25 \)
For a quadratic equation whose roots are \( \alpha \) and \( \beta \), the general form is:
\( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)
\( x^2 - (\alpha + \beta)x + (\alpha \beta) = 0 \)
Substitute the values we found for \( \alpha + \beta \) and \( \alpha \beta \):
\( x^2 - (16)x + (25) = 0 \)
\( x^2 - 16x + 25 = 0 \)
This is the required quadratic equation.
In simple words: We used the given average and geometric mean of the roots to find their sum and product. Then, we put these values into the standard formula for a quadratic equation, which uses the sum and product of its roots.

Exam Tip: Remember the relationships between roots and coefficients of a quadratic equation: for \( ax^2 + bx + c = 0 \), sum of roots \( (\alpha + \beta) = -b/a \) and product of roots \( (\alpha \beta) = c/a \). A direct application is \( x^2 - (\alpha+\beta)x + \alpha\beta = 0 \).

Question 32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation?
Answer: Let alpha and beta represent the roots of a quadratic equation. The arithmetic mean of alpha and beta is \( \frac{\alpha+\beta}{2} = 8 \). And the geometric mean of alpha and beta is \( \alpha\beta = 5^2 \). This results in \( \alpha + \beta = 16 \) and \( \alpha\beta = 25 \). The quadratic equation having alpha and beta as its roots is formed as: \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \) \( x^2 - 16x + 25 = 0 \).
In simple words: The average of the two roots is 8, and their product is 5 squared. Using these values, we can find the sum and product of the roots. Then, we put these numbers into the standard formula for a quadratic equation to get the final equation.

Exam Tip: Remember the standard form of a quadratic equation when roots are given: \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \). This formula is key for solving such problems quickly.