Given that the sum of some terms in a G.P is 315.

Let the number of terms be n.

We know that, sum of terms is

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

Given that the first term a is 5 and common ratio r is 2.

\(315=\dfrac{5(2^n-1)}{2-1}\)

2^{n}-1=63

2^{n }= 64 = (2)^{6}

n=6

Hence, the last term of the G.P = 6^{th} term = ar^{6 – 1} = (5)(2)^{5} = (5)(32) = 160

Therefore, the last term of the G.P. is 160.

Answered by Pragya Singh | 1 year 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 \( \dfrac{81}{2}\).