Sum of infinite number of terms in G.P. is 20 | vedclass

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Question

(b) $\frac{a}{{1 - r}} = 20 .....(i)$

$\frac{{{a^2}}}{{1 - {r^2}}} = 100 &hellip;..(ii)$

From $(i)$ and $(ii),$

$\frac{a}{{1 + r}} = 5$,&nbsp

Vedclass · Accepted Answer

$3/5$

Vedclass · Answer

(b) $\frac{a}{{1 - r}} = 20 .....(i)$

$\frac{{{a^2}}}{{1 - {r^2}}} = 100 &hellip;..(ii)$

From $(i)$ and $(ii),$

$\frac{a}{{1 + r}} = 5$, $\left[ {a = 20(1 - r)\,\,{\rm{by}}\,\,(i)} \right]$

==> $\frac{{20(1 - r)}}{{1 + r}} = 5$

$ \Rightarrow 5r = 3$

==> $r = 3/5$.

Sum of infinite number of terms in $G.P.$ is $20$ and sum of their square is $100$. The common ratio of $G.P.$ is

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