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$Given\;G.P\;is\;3,3^2,3^3,..........\\*Here\;first\;term,\;a=3\;and\;common\;ratio,\;r=3\\*Let\;sum\;of\;n\;terms\;of\;given\;G.P.\;will\;give\;120.\\*\Rightarrow \frac{a(r^n-1)}{r-1}=\frac{3(3^n-1)}{3-1}=120\\*\Rightarrow 3(3^n-1)=240 \\*\Rightarrow 3^n-1=80\\*\Rightarrow 3^n=81\Rightarrow 3^n=3^4\\* \therefore n=4\\*Hence,\;four\;terms\;required\;to\;give\;sum\;of\;120\;for\; given\;G.P.\\*$

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