The sum of first four terms of a GP is 3030 and that of last four terms is 960960. If the first and the last term of the GP is 2 and 512 respectively, find the common ratio.?

1 Answer
Jul 28, 2018

2root(3)2232.

Explanation:

Suppose that the common ratio (cr) of the GP in question is rr and n^(th)nth

term is the last term.

Given that, the first term of the GP is 22.

:."The GP is "{2,2r,2r^2,2r^3,..,2r^(n-4),2r^(n-3),2r^(n-2),2r^(n-1)}.

Given, 2+2r+2r^2+2r^3=30...(star^1), and,

2r^(n-4)+2r^(n-3)+2r^(n-2)+2r^(n-1)=960...(star^2).

We also know that the last term is 512.

:. r^(n-1)=512....................(star^3).

Now, (star^2) rArr r^(n-4)(2+2r+2r^2+2r^3)=960,

i.e., (r^(n-1))/r^3(2+2r+2r^2+2r^3)=960.

:. (512)/r^3(30)=960......[because, (star^1) & (star^3)].

:. r=root(3)(512*30/960)=2root(3)2, is the desired (real) cr!