How many terms of the G.P. 3,3/2,3/4 are needed to give the sum 765/128?

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How many terms of the G.P. 3,3/2,3/4 are needed to give the sum 765/128?

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3, $\frac{3}{2}$$\frac{3}{4}$

Here r = $\frac{1}{2}$< 1

sum of G.P = $\frac{a(1-r^n)}{1-r}$

⇒ $\frac{255}{128} = \frac{3(1-(\frac{1}{2})^n)}{1- (\frac{1}{2})}$ = $\frac{3(1- \frac{1}{2^n})}{\frac{1}{2}} = 6({1- \frac{1}{2^n}})$

⇒ $\frac{255}{256}$ = $1- \frac{1}{2^n}$

⇒ $\frac{1}{2^n} = 1 - \frac{255}{256}$ = $\frac{1}{256} = \frac{1}{2^8}$

⇒ n = 8