Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 .

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 $$\frac{98}{25}$$. 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 :

Accepted Answer

Let first term of G.P. be a and common ratio is r

Then, $${a \over {1 - r}} = 5$$ ...... (i)

$$a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}$$

$$\therefore$$ $${r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}$$

$$\therefore$$ Then, $${S_{21}} = {{21} \over 2}\left[ {2 \times 10ar + 20 \times 10a{r^2}} \right]$$

$$ = 21\left[ {10ar + 10\,.\,10a{r^2}} \right]$$

$$ = 21\,{a_{11}}$$

Solutions · 1