If the sum of first 6 term is 9 times to the | vedclass

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Question

(b) Under given conditions, we get

$ \Rightarrow $ $\frac{{a({r^6} - 1)}}{{(r - 1)}} = 9.\frac{{a({r^3} - 1)}}{{(r - 1)}}$ $(\because \;r > 1)$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b) Under given conditions, we get

$ \Rightarrow $ $\frac{{a({r^6} - 1)}}{{(r - 1)}} = 9.\frac{{a({r^3} - 1)}}{{(r - 1)}}$ $(\because \;r > 1)$

$ \Rightarrow $ ${r^6} - 1 = 9{r^3} - 9$

$ \Rightarrow $ ${({r^3})^2} - 9({r^3}) + 8 = 0$

$ \Rightarrow $ $({r^3} - 1)({r^3} - 8) = 0$

$ \Rightarrow $ $r = 1,\;\omega ,\;{\omega ^2}$

and $r = 2$. But $r = 1,\;\omega ,\;{\omega ^2}$ cannot satisfy the given condition.

Hence $r = 2$.

If the sum of first 6 term is $9$ times to the sum of first $3$ terms of the same $G.P.$, then the common ratio of the series will be

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