OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Exercise 9 (D)

Question 1. Find the sum of the terms of the indicated geometric sequence.
(i) 3, -6, 12,... to 6 terms
(ii) -2, -6, -18,... to 7 terms
(iii) \( \frac { 1 }{ 9 }, \frac { 1 }{ 3 }, 1, ... \) to 5 terms
(iv) 2, 1, \( \frac { 1 }{ 2 }, ... \) to 6 terms
(v) 1, \( \frac { 2 }{ 3 } \), \( \frac { 4 }{ 9 } \), ... to 10 terms
(vi) 0.15, 0.015, 0.0015, ... 20 terms
Answer:
(i) For the sequence 3, -6, 12,... to 6 terms:
The first term is \( a = 3 \).
The common ratio is \( r = \frac{-6}{3} = -2 \).
The number of terms is \( n = 6 \).
Since \( |r| = |-2| = 2 > 1 \), we use the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ S_6 = \frac{3((-2)^6-1)}{-2-1} \] \[ S_6 = \frac{3(64-1)}{-3} \] \[ S_6 = \frac{3(63)}{-3} \] \[ S_6 = -63 \] (ii) For the sequence -2, -6, -18,... to 7 terms:
The first term is \( a = -2 \).
The common ratio is \( r = \frac{-6}{-2} = 3 \).
The number of terms is \( n = 7 \).
Since \( |r| = |3| = 3 > 1 \), we use the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ S_7 = \frac{-2(3^7-1)}{3-1} \] \[ S_7 = \frac{-2(2187-1)}{2} \] \[ S_7 = -(2186) \] \[ S_7 = -2186 \] (iii) For the sequence \( \frac { 1 }{ 9 }, \frac { 1 }{ 3 }, 1, ... \) to 5 terms:
The first term is \( a = \frac{1}{9} \).
The common ratio is \( r = \frac{1/3}{1/9} = \frac{1}{3} \times 9 = 3 \).
The number of terms is \( n = 5 \).
Since \( |r| = |3| = 3 > 1 \), we use the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ S_5 = \frac{\frac{1}{9}(3^5-1)}{3-1} \] \[ S_5 = \frac{\frac{1}{9}(243-1)}{2} \] \[ S_5 = \frac{\frac{1}{9}(242)}{2} \] \[ S_5 = \frac{242}{18} \] \[ S_5 = \frac{121}{9} \] \[ S_5 = 13\frac{4}{9} \] (iv) For the sequence 2, 1, \( \frac { 1 }{ 2 }, ... \) to 6 terms:
The first term is \( a = 2 \).
The common ratio is \( r = \frac{1}{2} \).
The number of terms is \( n = 6 \).
Since \( |r| = |\frac{1}{2}| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_6 = \frac{2(1-(\frac{1}{2})^6)}{1-\frac{1}{2}} \] \[ S_6 = \frac{2(1-\frac{1}{64})}{\frac{1}{2}} \] \[ S_6 = 4(1-\frac{1}{64}) \] \[ S_6 = 4(\frac{64-1}{64}) \] \[ S_6 = 4(\frac{63}{64}) \] \[ S_6 = \frac{63}{16} \] \[ S_6 = 3\frac{15}{16} \] (v) For the sequence 1, \( \frac { 2 }{ 3 } \), \( \frac { 4 }{ 9 } \), ... to 10 terms:
The first term is \( a = 1 \).
The common ratio is \( r = \frac{2/3}{1} = \frac{2}{3} \).
The number of terms is \( n = 10 \).
Since \( |r| = |\frac{2}{3}| = \frac{2}{3} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_{10} = \frac{1(1-(\frac{2}{3})^{10})}{1-\frac{2}{3}} \] \[ S_{10} = \frac{1-(\frac{2}{3})^{10}}{\frac{1}{3}} \] \[ S_{10} = 3(1-(\frac{2}{3})^{10}) \] (vi) For the sequence 0.15, 0.015, 0.0015, ... 20 terms:
The first term is \( a = 0.15 \).
The common ratio is \( r = \frac{0.015}{0.15} = 0.1 \).
The number of terms is \( n = 20 \).
Since \( |r| = |0.1| = 0.1 < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_{20} = \frac{0.15(1-(0.1)^{20})}{1-0.1} \] \[ S_{20} = \frac{0.15(1-(0.1)^{20})}{0.9} \] \[ S_{20} = \frac{15}{90}(1-(0.1)^{20}) \] \[ S_{20} = \frac{1}{6}(1-(0.1)^{20}) \]In simple words: For each sequence, first find the starting number (a) and the constant multiplier (r). Then, use the correct formula for the sum of a geometric progression, choosing based on whether 'r' is bigger or smaller than 1. This helps us quickly add up all the terms without doing each addition separately.

🎯 Exam Tip: Always identify the first term (a), common ratio (r), and number of terms (n) correctly. Pay close attention to whether \( |r| < 1 \) or \( |r| > 1 \) to use the appropriate sum formula and avoid sign errors.

Question 2. Find the sum of the following series
(i) 12 + 6 + 3 + 1.5 + ... to 10 terms
(ii) 6-3 + 1\( \frac { 1 }{ 2 } \) – \( \frac { 3 }{ 4 } \)+... to 15 terms
(iii) 2 + 6 + 18 + 54 + ... to 12 terms
(iv) 6 + 12 + 24 + ... + 1536
Answer:
(i) For the series 12 + 6 + 3 + 1.5 + ... to 10 terms:
The first term is \( a = 12 \).
The common ratio is \( r = \frac{6}{12} = \frac{1}{2} \).
The number of terms is \( n = 10 \).
Since \( |r| = |\frac{1}{2}| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_{10} = \frac{12(1-(\frac{1}{2})^{10})}{1-\frac{1}{2}} \] \[ S_{10} = \frac{12(1-\frac{1}{1024})}{\frac{1}{2}} \] \[ S_{10} = 12 \times 2 (1-\frac{1}{1024}) \] \[ S_{10} = 24 (\frac{1024-1}{1024}) \] \[ S_{10} = 24 (\frac{1023}{1024}) \] (ii) For the series 6-3 + 1\( \frac { 1 }{ 2 } \) – \( \frac { 3 }{ 4 } \)+... to 15 terms:
The first term is \( a = 6 \).
The common ratio is \( r = \frac{-3}{6} = -\frac{1}{2} \).
The number of terms is \( n = 15 \).
Since \( |r| = |-\frac{1}{2}| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_{15} = \frac{6(1-(-\frac{1}{2})^{15})}{1-(-\frac{1}{2})} \] \[ S_{15} = \frac{6(1-(-\frac{1}{2})^{15})}{1+\frac{1}{2}} \] \[ S_{15} = \frac{6(1-(-\frac{1}{2})^{15})}{\frac{3}{2}} \] \[ S_{15} = 6 \times \frac{2}{3} (1-(-\frac{1}{2})^{15}) \] \[ S_{15} = 4(1+\frac{1}{2^{15}}) \] \[ S_{15} = 4(1+\frac{1}{32768}) \] \[ S_{15} = 4(\frac{32768+1}{32768}) \] \[ S_{15} = 4(\frac{32769}{32768}) \] \[ S_{15} = \frac{32769}{8192} \] (iii) For the series 2 + 6 + 18 + 54 + ... to 12 terms:
The first term is \( a = 2 \).
The common ratio is \( r = \frac{6}{2} = 3 \).
The number of terms is \( n = 12 \).
Since \( |r| = |3| = 3 > 1 \), we use the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ S_{12} = \frac{2(3^{12}-1)}{3-1} \] \[ S_{12} = \frac{2(3^{12}-1)}{2} \] \[ S_{12} = 3^{12}-1 \] (iv) For the series 6 + 12 + 24 + ... + 1536:
The first term is \( a = 6 \).
The common ratio is \( r = \frac{12}{6} = 2 \).
The last term is \( l = 1536 \).
We use the formula for the nth term \( l = ar^{n-1} \) to find \( n \).
\( 1536 = 6(2^{n-1}) \)
\( \implies \frac{1536}{6} = 2^{n-1} \)
\( \implies 256 = 2^{n-1} \)
\( \implies 2^8 = 2^{n-1} \)
\( \implies n-1 = 8 \)
\( \implies n = 9 \)
Now, we find the sum using \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ S_9 = \frac{6(2^9-1)}{2-1} \] \[ S_9 = \frac{6(512-1)}{1} \] \[ S_9 = 6 \times 511 \] \[ S_9 = 3066 \]In simple words: For each problem, first check if the series is geometric and find its first term and common ratio. If you need to find the sum up to a specific term, use the sum formula. If the last term is given instead of the number of terms, first use the nth term formula to find how many terms there are, and then calculate the total sum.

🎯 Exam Tip: When the last term is given, remember to first calculate 'n' using the formula \( a_n = ar^{n-1} \) before using the sum formula. This is a common multi-step problem type.

Question 3. Find the sum to n terms of the following series.
(i) 12 + 6 + 3 + 1 \( \frac { 1 }{ 2 } \) + ... n terms
(ii) 20 – 10 + 5 – 2 \( \frac { 1 }{ 2 } \) + ... n terms
(iii) 9 – 3, +1 – \( \frac { 1 }{ 3 } \) +... n terms
(iv) \( \sqrt{3} \) + 3 + 3\( \sqrt{3} \) + 9 + ... n terms
(v) 0.9 + 0.09 + 0.009 + 0.0009 + ...
Answer:
(i) For the series 12 + 6 + 3 + 1 \( \frac { 1 }{ 2 } \) + ... n terms:
The first term is \( a = 12 \).
The common ratio is \( r = \frac{6}{12} = \frac{1}{2} \).
Since \( |r| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_n = \frac{12(1-(\frac{1}{2})^n)}{1-\frac{1}{2}} \] \[ S_n = \frac{12(1-(\frac{1}{2})^n)}{\frac{1}{2}} \] \[ S_n = 12 \times 2 (1-(\frac{1}{2})^n) \] \[ S_n = 24(1-(\frac{1}{2})^n) \] (ii) For the series 20 – 10 + 5 – 2 \( \frac { 1 }{ 2 } \) + ... n terms:
The first term is \( a = 20 \).
The common ratio is \( r = \frac{-10}{20} = -\frac{1}{2} \).
Since \( |r| = |-\frac{1}{2}| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_n = \frac{20(1-(-\frac{1}{2})^n)}{1-(-\frac{1}{2})} \] \[ S_n = \frac{20(1-(-\frac{1}{2})^n)}{1+\frac{1}{2}} \] \[ S_n = \frac{20(1-(-\frac{1}{2})^n)}{\frac{3}{2}} \] \[ S_n = 20 \times \frac{2}{3} (1-(-\frac{1}{2})^n) \] \[ S_n = \frac{40}{3}(1-(-\frac{1}{2})^n) \] (iii) For the series 9 – 3, +1 – \( \frac { 1 }{ 3 } \)+... n terms:
The first term is \( a = 9 \).
The common ratio is \( r = \frac{-3}{9} = -\frac{1}{3} \).
Since \( |r| = |-\frac{1}{3}| = \frac{1}{3} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_n = \frac{9(1-(-\frac{1}{3})^n)}{1-(-\frac{1}{3})} \] \[ S_n = \frac{9(1-(-\frac{1}{3})^n)}{1+\frac{1}{3}} \] \[ S_n = \frac{9(1-(-\frac{1}{3})^n)}{\frac{4}{3}} \] \[ S_n = 9 \times \frac{3}{4} (1-(-\frac{1}{3})^n) \] \[ S_n = \frac{27}{4}(1-(-\frac{1}{3})^n) \] (iv) For the series \( \sqrt{3} \) + 3 + 3\( \sqrt{3} \) + 9 + ... n terms:
The first term is \( a = \sqrt{3} \).
The common ratio is \( r = \frac{3}{\sqrt{3}} = \sqrt{3} \).
Since \( |r| = |\sqrt{3}| \approx 1.732 > 1 \), we use the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ S_n = \frac{\sqrt{3}((\sqrt{3})^n-1)}{\sqrt{3}-1} \] To simplify, multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{3}+1 \).
\[ S_n = \frac{\sqrt{3}((\sqrt{3})^n-1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \] \[ S_n = \frac{\sqrt{3}(\sqrt{3}^n-1)(\sqrt{3}+1)}{3-1} \] \[ S_n = \frac{\sqrt{3}(\sqrt{3}^n-1)(\sqrt{3}+1)}{2} \] \[ S_n = \frac{3+\sqrt{3}}{2}((\sqrt{3})^n-1) \] (v) For the series 0.9 + 0.09 + 0.009 + 0.0009 + ...:
The first term is \( a = 0.9 \).
The common ratio is \( r = \frac{0.09}{0.9} = 0.1 \).
Since \( |r| = |0.1| = 0.1 < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ S_n = \frac{0.9(1-(0.1)^n)}{1-0.1} \] \[ S_n = \frac{0.9(1-(0.1)^n)}{0.9} \] \[ S_n = 1-(0.1)^n \]In simple words: To find the sum of a geometric series up to 'n' terms, you first need to identify the first term (a) and the common ratio (r). Then, pick the right sum formula based on whether 'r' is less than 1 or greater than 1. Make sure to keep 'n' as a variable in your final answer.

🎯 Exam Tip: Always state the formula you are using for the sum of a geometric series. If the common ratio is negative, be careful with the signs when calculating \( r^n \), especially when 'n' is odd or even.

Question 4. Use the formula \( S_n = \frac{a(r^n-1)}{r-1} \) to find:
(i) \( a_1 \) and \( S_n \) if \( a_n = 1000, r = 10 \) and \( n = 7 \)
(ii) n and \( S_n \) if \( a_n = 5, a_1 = 320, r = 2 \)
(iii) \( a_9 \) and \( a_1 \) if \( n = 9, r = 2, S_n = 1022 \)
Answer:
(i) Given: \( a_n = 1000, r = 10, n = 7 \). We need to find \( a_1 \) (first term) and \( S_n \) (sum of terms).
First, use the formula for the nth term: \( a_n = a_1 r^{n-1} \).
\( 1000 = a_1 \times 10^{7-1} \)
\( \implies 1000 = a_1 \times 10^6 \)
\( \implies a_1 = \frac{1000}{10^6} \)
\( \implies a_1 = \frac{10^3}{10^6} \)
\( \implies a_1 = 10^{-3} \)
\( \implies a_1 = 0.001 \)
Now, find \( S_n \) using the given formula \( S_n = \frac{a_1(r^n-1)}{r-1} \).
\[ S_7 = \frac{0.001(10^7-1)}{10-1} \] \[ S_7 = \frac{0.001(10000000-1)}{9} \] \[ S_7 = \frac{0.001 \times 9999999}{9} \] \[ S_7 = \frac{9999.999}{9} \] \[ S_7 = 1111.111 \] (ii) Given: \( a_1 = 5, a_n = 320, r = 2 \). We need to find n (number of terms) and \( S_n \) (sum of terms).
First, use the formula for the nth term: \( a_n = a_1 r^{n-1} \).
\( 320 = 5 \times 2^{n-1} \)
\( \implies \frac{320}{5} = 2^{n-1} \)
\( \implies 64 = 2^{n-1} \)
\( \implies 2^6 = 2^{n-1} \)
\( \implies n-1 = 6 \)
\( \implies n = 7 \)
Now, find \( S_n \) using the formula \( S_n = \frac{a_1(r^n-1)}{r-1} \).
\[ S_7 = \frac{5(2^7-1)}{2-1} \] \[ S_7 = \frac{5(128-1)}{1} \] \[ S_7 = 5 \times 127 \] \[ S_7 = 635 \] (iii) Given: \( n = 9, r = 2, S_n = 1022 \). We need to find \( a_9 \) (9th term) and \( a_1 \) (first term).
First, use the formula for the sum of terms: \( S_n = \frac{a_1(r^n-1)}{r-1} \).
\( 1022 = \frac{a_1(2^9-1)}{2-1} \)
\( \implies 1022 = \frac{a_1(512-1)}{1} \)
\( \implies 1022 = a_1 \times 511 \)
\( \implies a_1 = \frac{1022}{511} \)
\( \implies a_1 = 2 \)
Now, find \( a_9 \) using the formula for the nth term: \( a_n = a_1 r^{n-1} \).
\( a_9 = 2 \times 2^{9-1} \)
\( \implies a_9 = 2 \times 2^8 \)
\( \implies a_9 = 2 \times 256 \)
\( \implies a_9 = 512 \)
In simple words: When you have some information about a geometric progression, like its last term, first term, common ratio, or sum, you can use the formulas for the nth term \( (a_n = a_1 r^{n-1}) \) and sum of n terms \( (S_n = \frac{a_1(r^n-1)}{r-1}) \) to find the missing parts. You might need to use both formulas in steps.

🎯 Exam Tip: Remember to solve for the unknown in sequence. Often, you'll need to find 'n' or 'a' using one formula before you can calculate the sum or a specific term with the other.

Question 5. If \( \{a_n\} \) is a G.S. (i.e., geometric sequence) and \( a_1 = 4, r = 5 \), find \( a_6 \) and \( S_6 \).
Answer:
Given a geometric sequence with first term \( a_1 = 4 \) and common ratio \( r = 5 \).
We need to find the 6th term \( (a_6) \) and the sum of the first 6 terms \( (S_6) \).
First, find \( a_6 \) using the formula \( a_n = a_1 r^{n-1} \):
\( a_6 = 4 \times 5^{6-1} \)
\( \implies a_6 = 4 \times 5^5 \)
\( \implies a_6 = 4 \times (5 \times 5 \times 5 \times 5 \times 5) \)
\( \implies a_6 = 4 \times 3125 \)
\( \implies a_6 = 12500 \)
Next, find \( S_6 \) using the sum formula \( S_n = \frac{a_1(r^n-1)}{r-1} \) (since \( r=5 > 1 \)):
\[ S_6 = \frac{4(5^6-1)}{5-1} \] \[ S_6 = \frac{4(15625-1)}{4} \] \[ S_6 = 15624 \]In simple words: To find a specific term in a geometric sequence, multiply the first term by the common ratio raised to one less than the term number. To find the sum of terms, use the sum formula, making sure to apply the ratio and number of terms correctly.

🎯 Exam Tip: Always remember that the exponent for 'r' in the nth term formula \( a_n = ar^{n-1} \) is \( n-1 \), but in the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \), the exponent is 'n'.

Question 6. How many terms of the G.P. 3, \( \frac { 3 }{ 2 } \), \( \frac { 3 }{ 4 } \),...are needed to give the sum \( \frac { 3069 }{ 512 } \)?
Answer:
Given the G.P.: 3, \( \frac { 3 }{ 2 } \), \( \frac { 3 }{ 4 } \), ...
The first term is \( a = 3 \).
The common ratio is \( r = \frac{3/2}{3} = \frac{1}{2} \).
The sum of the terms is \( S_n = \frac{3069}{512} \).
Since \( |r| = \frac{1}{2} < 1 \), we use the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \).
\[ \frac{3069}{512} = \frac{3(1-(\frac{1}{2})^n)}{1-\frac{1}{2}} \] \[ \frac{3069}{512} = \frac{3(1-(\frac{1}{2})^n)}{\frac{1}{2}} \] \[ \frac{3069}{512} = 3 \times 2 (1-(\frac{1}{2})^n) \] \[ \frac{3069}{512} = 6 (1-(\frac{1}{2})^n) \]
\( \implies \frac{3069}{512 \times 6} = 1-(\frac{1}{2})^n \)
\( \implies \frac{3069}{3072} = 1-(\frac{1}{2})^n \)
\( \implies (\frac{1}{2})^n = 1 - \frac{3069}{3072} \)
\( \implies (\frac{1}{2})^n = \frac{3072-3069}{3072} \)
\( \implies (\frac{1}{2})^n = \frac{3}{3072} \)
\( \implies (\frac{1}{2})^n = \frac{1}{1024} \)
\( \implies (\frac{1}{2})^n = (\frac{1}{2})^{10} \)
\( \implies n = 10 \)
Therefore, 10 terms are needed to give the sum \( \frac{3069}{512} \).In simple words: To find out how many terms sum up to a certain value in a geometric progression, first identify the starting number and the common ratio. Then, set up the sum formula with the given sum and solve for 'n' by matching the bases of the powers.

🎯 Exam Tip: When solving for 'n' in geometric progression problems, simplify the equation to have terms with the same base on both sides. This makes comparing the exponents straightforward.

Question 7. The sum of some terms of a G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Answer:
Given: Sum of terms \( S_n = 315 \).
First term \( a_1 = 5 \).
Common ratio \( r = 2 \).
We need to find the number of terms (n) and the last term \( (a_n) \).
Since \( r = 2 > 1 \), we use the sum formula \( S_n = \frac{a_1(r^n-1)}{r-1} \).
\( 315 = \frac{5(2^n-1)}{2-1} \)
\( \implies 315 = \frac{5(2^n-1)}{1} \)
\( \implies 315 = 5(2^n-1) \)
\( \implies \frac{315}{5} = 2^n-1 \)
\( \implies 63 = 2^n-1 \)
\( \implies 63+1 = 2^n \)
\( \implies 64 = 2^n \)
\( \implies 2^6 = 2^n \)
\( \implies n = 6 \)
Now, find the last term \( a_n \) using the formula \( a_n = a_1 r^{n-1} \):
\( a_6 = 5 \times 2^{6-1} \)
\( \implies a_6 = 5 \times 2^5 \)
\( \implies a_6 = 5 \times 32 \)
\( \implies a_6 = 160 \)
The number of terms is 6 and the last term is 160.In simple words: If you know the sum, first term, and common ratio of a geometric sequence, you can find the number of terms using the sum formula. Once you have the number of terms, you can then easily find the last term using the formula for the nth term.

🎯 Exam Tip: When given the sum, first term, and common ratio, prioritize finding 'n' (number of terms) using the sum formula. This 'n' will then be used to calculate the last term.

Question 8. Given a G.P. with \( a = 729 \) and 7th term \( = 64 \), determine \( S_7 \).
Answer:
Given: First term \( a = 729 \).
7th term \( a_7 = 64 \).
We need to determine the sum of the first 7 terms \( (S_7) \).
First, use the formula for the nth term \( a_n = ar^{n-1} \) to find the common ratio \( r \).
\( a_7 = ar^{7-1} \)
\( \implies 64 = 729 r^6 \)
\( \implies r^6 = \frac{64}{729} \)
\( \implies r^6 = (\frac{2}{3})^6 \)
\( \implies r = \frac{2}{3} \)
Now, find \( S_7 \) using the sum formula \( S_n = \frac{a(1-r^n)}{1-r} \) (since \( r=\frac{2}{3} < 1 \)):
\[ S_7 = \frac{729(1-(\frac{2}{3})^7)}{1-\frac{2}{3}} \] \[ S_7 = \frac{729(1-\frac{128}{2187})}{\frac{1}{3}} \] \[ S_7 = 729 \times 3 (1-\frac{128}{2187}) \] \[ S_7 = 2187 (\frac{2187-128}{2187}) \] \[ S_7 = 2187 (\frac{2059}{2187}) \] \[ S_7 = 2059 \]In simple words: When you know the first term and a later term in a geometric sequence, first find the common ratio by using the nth term formula. Once you have the common ratio, you can easily calculate the sum of the terms using the sum formula.

🎯 Exam Tip: If the common ratio is a fraction, be careful with its powers, especially when determining if \( |r| \) is greater or less than 1. This helps in choosing the correct sum formula.

Question 9. Find the sum of the series 2 + 6 + 18 + ... + 4374.
Answer:
Given the series: 2 + 6 + 18 + ... + 4374.
The first term is \( a = 2 \).
The common ratio is \( r = \frac{6}{2} = 3 \).
The last term is \( l = 4374 \).
First, find the number of terms (n) using the formula for the nth term \( l = ar^{n-1} \).
\( 4374 = 2 \times 3^{n-1} \)
\( \implies \frac{4374}{2} = 3^{n-1} \)
\( \implies 2187 = 3^{n-1} \)
\( \implies 3^7 = 3^{n-1} \)
\( \implies n-1 = 7 \)
\( \implies n = 8 \)
Now, find the sum of the 8 terms using the formula \( S_n = \frac{a(r^n-1)}{r-1} \) (since \( r=3 > 1 \)):
\[ S_8 = \frac{2(3^8-1)}{3-1} \] \[ S_8 = \frac{2(6561-1)}{2} \] \[ S_8 = 6560 \]In simple words: When a geometric series has a known first and last term, start by finding how many terms are in the series using the nth term formula. After you find the number of terms, use the sum formula to add them all up.

🎯 Exam Tip: Recognizing powers of numbers (like \( 2187 = 3^7 \)) can save time when solving for 'n' by comparing exponents.

Question 10. How many terms of the sequence \( \sqrt{3}, 3, 3\sqrt{3}, ... \) must be taken to make the sum \( 39 + 13\sqrt{3} \)?
Answer:
Given the sequence: \( \sqrt{3}, 3, 3\sqrt{3}, ... \)
The first term is \( a = \sqrt{3} \).
The common ratio is \( r = \frac{3}{\sqrt{3}} = \sqrt{3} \).
The sum of the terms is \( S_n = 39 + 13\sqrt{3} \).
We need to find the number of terms (n).
Since \( r = \sqrt{3} > 1 \), we use the sum formula \( S_n = \frac{a(r^n-1)}{r-1} \).
\[ 39 + 13\sqrt{3} = \frac{\sqrt{3}((\sqrt{3})^n-1)}{\sqrt{3}-1} \] Multiply both sides by \( (\sqrt{3}-1) \):
\( (39 + 13\sqrt{3})(\sqrt{3}-1) = \sqrt{3}((\sqrt{3})^n-1) \)
\( \implies 39\sqrt{3} - 39 + 13\sqrt{3}\sqrt{3} - 13\sqrt{3} = \sqrt{3}((\sqrt{3})^n-1) \)
\( \implies 39\sqrt{3} - 39 + 13 \times 3 - 13\sqrt{3} = \sqrt{3}((\sqrt{3})^n-1) \)
\( \implies 39\sqrt{3} - 39 + 39 - 13\sqrt{3} = \sqrt{3}((\sqrt{3})^n-1) \)
\( \implies 26\sqrt{3} = \sqrt{3}((\sqrt{3})^n-1) \) Divide both sides by \( \sqrt{3} \):
\( 26 = (\sqrt{3})^n-1 \)
\( \implies 26+1 = (\sqrt{3})^n \)
\( \implies 27 = (\sqrt{3})^n \) We know that \( 27 = 3^3 \). Also, \( \sqrt{3} \) can be written as \( 3^{1/2} \).
So, \( 3^3 = (3^{1/2})^n \)
\( \implies 3^3 = 3^{n/2} \)
\( \implies 3 = \frac{n}{2} \)
\( \implies n = 6 \)
Therefore, 6 terms must be taken to make the sum \( 39 + 13\sqrt{3} \).In simple words: To find how many terms sum up to a given value, set up the sum formula for a geometric progression. Simplify the equation by multiplying by the conjugate of the denominator, then solve for 'n' by converting all terms to the same base and comparing the exponents.

🎯 Exam Tip: When dealing with square roots in the common ratio or terms, remember to rationalize denominators when needed. Also, express all numbers as powers of the same base (e.g., 27 as \( 3^3 \) and \( \sqrt{3} \) as \( 3^{1/2} \)) to easily solve for 'n'.