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 9825then the sum of the first 21 terms of an A.P Whose first term is 10 ar, nth term is an and the common difference is 10 ar2 is

strong{font-weight:bold !important} S∞=5⇒a1−r=5→(1) S5=9825⇒a1−r51−r=9825 →(2)(1)÷(2)⇒ a1−r⋅1−ra1−r5=5×2598 ⇒11−r5=12598⇒1−r5=98125∴ r5=1−98125=125−98125=27125⇒r=3535 S21=2122.10ar+20.10ar2S21=21220ar+200ar2 =2110ar+100ar2=21⋅a11

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 9825then the sum of the first 21 terms of an A.P Whose first term is 10 ar, nth term is an and the common difference is 10 ar2 is

strong{font-weight:bold !important} S∞=5⇒a1−r=5→(1) S5=9825⇒a1−r51−r=9825 →(2)(1)÷(2)⇒ a1−r⋅1−ra1−r5=5×2598 ⇒11−r5=12598⇒1−r5=98125∴ r5=1−98125=125−98125=27125⇒r=3535 S21=2122.10ar+20.10ar2S21=21220ar+200ar2 =2110ar+100ar2=21⋅a11

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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

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