Question

How many terms of G.P. \(3,3^2,3^3\), … are needed to give the sum 120?

Answer 1

The given G.P. is \(3,3^2,3^3,\) …

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

Sn = \(a\frac{(r^n-1)}{r-1}\)

Here, a = 3 and r = 3

∴ Sn = 120 = \(3\frac{(3^n-1)}{3-1}\)

⇒ 120 = \(3\frac{(3^n-1)}{2}\)

⇒ \(\frac{120\times2}{3}=3^n-1\)

⇒ \(3^n-1\) = 80

⇒ \(3^n\) = 80

⇒ \(3^n\) =\(3^4\)

∴ n = 4

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