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

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Solution

The given G.P. is 3, 3², 3³, …

Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3

∴ n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

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Similar questions

3, 3^{2}, 3^{3}, . . .

120

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

\frac{3}{2}, \frac{3}{4}

\frac{3069}{512}

3, \frac{3}{2}, \frac{3}{4} . . .,

\frac{3069}{512} ?

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