If the sum of n terms of a G.P. is 255,the n^ | vedclass

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(A) Given that the sum of $n$ terms $S_n = \frac{a(r^n - 1)}{r - 1} = 255$ (since $r > 1$) .....$(i)$The $n^{th}$ term $a_n = ar^{n-1} = 128$ .....$(i

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(A) Given that the sum of $n$ terms $S_n = \frac{a(r^n - 1)}{r - 1} = 255$ (since $r > 1$) .....$(i)$The $n^{th}$ term $a_n = ar^{n-1} = 128$ .....$(ii)$Common ratio $r = 2$ .....$(iii)$From $(ii)$ and $(iii)$,we have $a(2^{n-1}) = 128$ .....$(iv)$From $(i)$ and $(iii)$,we have $\frac{a(2^n - 1)}{2 - 1} = 255 \Rightarrow a(2^n - 1) = 255$ .....$(v)$Dividing $(v)$ by $(iv)$:$\frac{a(2^n - 1)}{a(2^{n-1})} = \frac{255}{128}$$\frac{2^n - 1}{2^{n-1}} = \frac{255}{128}$$\frac{2^n}{2^{n-1}} - \frac{1}{2^{n-1}} = \frac{255}{128}$$2 - \frac{1}{2^{n-1}} = \frac{255}{128}$$\frac{1}{2^{n-1}} = 2 - \frac{255}{128} = \frac{256 - 255}{128} = \frac{1}{128}$$2^{n-1} = 128 = 2^7$$n - 1 = 7 \Rightarrow n = 8$Substituting $n = 8$ into $(iv)$:$a(2^{8-1}) = 128$$a(2^7) = 128$$a(128) = 128$$a = 1$.

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

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