If the sum of n terms of a G.P. is 255 and {n | vedclass

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Question

(a) Given that $\frac{{a({r^n} - 1)}}{{r - 1}} = 255$ $(\because \;\;r > 1)$ &hellip;..$(i)$

$a{r^{n - 1}} = 128$ &hellip;..$(ii)$

and common

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$1$

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(a) Given that $\frac{{a({r^n} - 1)}}{{r - 1}} = 255$ $(\because \;\;r > 1)$ &hellip;..$(i)$

$a{r^{n - 1}} = 128$ &hellip;..$(ii)$

and common ratio $r = 2$ &hellip;..$(iii)$

From $(iii), (i)$ and $(ii)$

we get $a{2^{n - 1}} = 128$  &hellip;..$(iv)$

and $\frac{{a({2^n} - 1)}}{{2 - 1}} = 255$ .....$(v)$

Dividing $(v)$ by $(iv)$

we get $\frac{{{2^n} - 1}}{{{2^{n - 1}}}} = \frac{{255}}{{128}}$

$ \Rightarrow $$2 - {2^{ - n + 1}} = \frac{{255}}{{128}}$

$ \Rightarrow $${2^{ - n}} = {2^{ - 8}}$

$ \Rightarrow $$n = 8$

Putting $n = 8$ in equation $(iv),$

we have $a\;.\;{2^7} = 128 = {2^7}$or $a = 1$.

If the sum of $n$ terms of a $G.P.$ is $255$ and ${n^{th}}$ terms is $128$ and common ratio is $2$, then first term will be

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