Given G.P. is 3, 3^{2}, 3^{3}, …

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

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

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

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

\( 3^n-1=80\)

3^{n}=81

3^{n}=3^{4}

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 | 2 years agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).