Question

Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5. Let the sum of its first five terms be 98/25. Then the sum of the first 21 terms of an AP, whose first term is 10ar, n^th term is a_n and the common difference is 10ar², is equal to

• A. 21 a₁₁
• B. 22 a₁₁
• C. 15 a₁₆
• D. 14 a₁₆

Answer 1

The Correct Option is A

To solve this problem, we must work through several steps dealing with geometric progressions (G.P.) and arithmetic progressions (A.P.). Let’s break it down step-by-step:

1. We are given the sum of an infinite G.P. with the first term \(a\) and a common ratio \(r\). For an infinite G.P., the sum is given by: \(S_\infty = \frac{a}{1 - r} = 5\). 
   From this, we have \(a = 5(1 - r)\).
2. Next, we are given that the sum of the first five terms of this G.P. is \(\frac{98}{25}\). The formula for the sum of the first \(n\) terms of a G.P. is: \(S_n = \frac{a(1 - r^n)}{1 - r}\). 
   Substituting for the first five terms: \(\frac{a(1 - r^5)}{1 - r} = \frac{98}{25}\).
3. By substituting \(a = 5(1 - r)\) from the first equation into the second, the equation becomes: \(\frac{5(1 - r)(1 - r^5)}{1 - r} = \frac{98}{25}\). This simplifies to: \(5(1 - r^5) = \frac{98}{25}\).
4. Solving for \(r\), we find: \((1 - r^5) = \frac{98}{125}\) 
   This implies: \(r^5 = \frac{27}{125}\). 
   Therefore, \(r = \left(\frac{27}{125}\right)^{\frac{1}{5}} = \frac{3}{5}\).
5. Substitute back to find \(a\): \(a = 5\left(1 - \frac{3}{5}\right) = 2\).
6. Now, for the A.P., we are told that the first term \(A_1 = 10ar\). 
   With \(a = 2\) and \(r = \frac{3}{5}\), we have: \(A_1 = 10 \times 2 \times \frac{3}{5} = 12\).
7. The common difference is \(d = 10ar^2\): \(d = 10 \times 2 \times \left(\frac{3}{5}\right)^2 = \frac{36}{5}\).
8. The sum of the first \(21\) terms of an A.P. is given by: \(S_{21} = \frac{21}{2} \times \left(2 \times A_1 + (21 - 1) \times d\right)\). Substituting the known values: \(S_{21} = \frac{21}{2} \times (2 \times 12 + 20 \times \frac{36}{5})\).
9. Simplifying, we find: \(S_{21} = \frac{21}{2} \times (24 + 144) = \frac{21}{2} \times 168 = 21 \times 84 = 1764\).
10. Next, calculate \(a_{11}\) which is the 11th term of the A.P.: \(a_{11} = A_1 + 10d = 12 + 10 \times \frac{36}{5} = 12 + 72 = 84\).
11. So, the sum \(1764\) can be expressed as: \(1764 = 21 \times 84 = 21 \times a_{11}\). Therefore, the correct answer is \(21 \ a_{11}\).