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 $\frac{98}{25}$ then the sum of the first 21 terms of an A.P Whose first term is 10 ar, n^thterm is a_n and the common difference is 10 ar² is

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$22 a_{11}$

$15 a_{16}$

$21 a_{11}$

$14 a_{16}$

answer is A.

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Detailed Solution

$\begin{array}{l} S_{\infty} = 5 \Rightarrow \frac{a}{1 - r} = 5 \to (1) S_{5} = \frac{98}{25} \Rightarrow \frac{a (1 - r^{5})}{1 - r} = \frac{98}{25} \to (2) \\ (1) \div (2) \Rightarrow \frac{a}{1 - r} \cdot \frac{1 - r}{a (1 - r^{5})} = 5 \times \frac{25}{98} \Rightarrow \frac{1}{1 - r^{5}} = \frac{125}{98} \Rightarrow 1 - r^{5} = \frac{98}{125} \\ ∴ r^{5} = 1 - \frac{98}{125} = \frac{125 - 98}{125} = \frac{27}{125} \Rightarrow r = {(\frac{3}{5})}^{\frac{3}{5}} S_{21} = \frac{21}{2} [2 .10 ar + 20 .10 {ar}^{2}] \\ S_{21} = \frac{21}{2} [20 ar + 200 {ar}^{2}] = 21 [10 ar + 100 {ar}^{2}] = 21 \cdot a_{11} \end{array}$