Sum of first four terms of a geometric series is #15# and next four terms is #240#. Find the first term and common ratio of the series?

1 Answer
Mar 28, 2017

There are two possibilities i.e. either #a_1=1# and #r=2#

or #a_1=-3# and #r=-2#.

Explanation:

In a geometric sequence, whose first term is #a_1# and common ratio is #r#, while #n^(th)# term #a_n=a_ar^((n-1))#, sum of first #n# terms is given by #S_n=(a_1(r^n-1))/(r-1)#.

Here assume that first term is #a_1# and common ratio is #r#. As sum of first four terms is #15#

#S_4=(a_1(r^4-1))/(r-1)=15# .........................(A)

i.e. #a_1+a_1r+a_1r^2+a_1r^3=15#

i.e. #a_1(1+r+r^2+r^3)=15# .........................(1)

and #a_1(r^4+r^5+r^6+r^7)=240#

i.e. #a_1r^4(1+r+r^2+r^3)=240# .........................(2)

Dividing (2) be (1), we get

#r^4=240/15=16#

i.e. #r=+-2#

If #r=2#, #a_1=15/(1+2+4+8)=1#

and if #r=-2#, #a_1=15/(1-2+4-8)=15/(-5)=-3#.

Hence sequence is

either #{1,2,4,8,16,32,64,18}# i.e. #a_1=1# and #r=2#

or #{-3,+6,-12,+24,-48,96,-192,384}# i.e. #a_1=-3# and #r=-2#