How many terms of G.P. 3,32,33,˙˙˙are needed to give the sum 120?

Solution: In the given GP,a=3,r=3Sn=120∴ar−1rn−1=120⇒33−1rn−1=120⇒3n−1=80⇒3n=81⇒3n=34n=4

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How many terms of G.P. {3},{3}^{{2}},{3}^{{3}},{{{}}}are neede

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How many terms of G.P. $3, 3^{2}, 3^{3}, \dot{\dot{˙}}$ are needed to give the sum 120?

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How many terms of G.P.

3, 3^{2}, 3^{3}, \dot{\dot{˙}}

are needed to give the sum 120?

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Solution: In the given GP,

a = 3, r = 3

S_{n} = 120

∴ a \frac{r ^{n} - 1}{r - 1} = 120

\Rightarrow 3 \frac{r ^{n} - 1}{3 - 1} = 120

\Rightarrow 3^{n} - 1 = 80 \Rightarrow 3^{n} = 81

\Rightarrow 3^{n} = 3^{4}

n = 4

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Question Text	How many terms of G.P. $3, 3^{2}, 3^{3}, \dot{\dot{˙}}$ are needed to give the sum 120?
Answer Type	Text solution:1
Upvotes	150