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

Asked by Pragya Singh | 1 year ago |  97

##### Solution :-

Given G.P. is 3, 32, 33, …

Let’s consider that n terms of this G.P. be required to obtain the sum of 120.

We know that,

$$s_n=\dfrac{a(1-r^n)}{1-r}$$

Here, a = 3 and r = 3

$$s_n=120=\dfrac{3(3^n-1)}{3-1}$$

$$s_n=120=\dfrac{3(3^n-1)}{2}$$

$$\dfrac{120\times 2}{3}=3^n-1$$

$$3^n-1=80$$

3n=81

3n=34

Equating the exponents we get, n = 4

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

Answered by Abhisek | 1 year ago

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