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 a 11
• B. 22 a 11
• C. 15 a 16
• D. 14 a 16

Answer 1

Answer: (A)

21 a 11

Answer 2

The correct answer is (A) : 21 a 11
Let first term of G.P. be a and common ratio is r
Then, $\frac{α}{1-r} = 5....(i)$
$α =\frac{(r^5-1)}{(r-1)}$
$= \frac{98}{25}$
$⇒ 1-r^5 = \frac{98}{125}$
$∴ r^5 = \frac{27}{125}$
$r = (\frac{3}{5})^{\frac{3}{5}}$
∴ Then, $S_{21} = \frac{21}{2}[2×10ar + 20×10ar^2]$
$= 21[10ar+10.10ar^2]$
$= ^{21}α_{11}$