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 AP, whose first term is $10\mathrm{a r}, \mathrm{n}^{\text {th }}$ term is $\mathrm{a}_{\mathrm{n}}$ and the common difference is $10 \mathrm{ar}^{2}$, is equal to :
Solution
<p>Let first term of G.P. be a and common ratio is r</p>
<p>Then, ${a \over {1 - r}} = 5$ ...... (i)</p>
<p>$$a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}$$</p>
<p>$\therefore$ ${r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}$</p>
<p>$\therefore$ Then, ${S_{21}} = {{21} \over 2}\left[ {2 \times 10ar + 20 \times 10a{r^2}} \right]$</p>
<p>$= 21\left[ {10ar + 10\,.\,10a{r^2}} \right]$</p>
<p>$= 21\,{a_{11}}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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