The sum of first four terms of a GP is #30# and that of last four terms is #960#. If the first and the last term of the GP is 2 and 512 respectively, find the common ratio.?

Question

3691 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-28T07:41:27

# 2root(3)2#.

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

term is the last term.

Given that, the first term of the GP is #2#.

#:."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!