The 3rd,6th,and 12th terms of an Arithmetic progression are successive terms of a Geometric progression.How to show that the 4th,8th,and 16th terms of the A.P are also successive terms of a G.P?

1 Answer
Jul 31, 2017

Please see below.

Explanation:

Let the first term of Arithmetic Progression (AP) be #a# and common difference be #d#.

Assume #d# is non-zero because if #d=0# then all terms of AP are same and #3^(rd)#, #6^(th)# and #12^(th)# too will be in Geometric Progression (GP) and so are #4^(th)#, #8^(th)# and #16^(th)# terms.

Asssuming #a!=d# and #3^(rd)#, #6^(th)# and #12^(th)# terms are in GP, we have

#(a+5d)^2=(a+2d)(a+11d)#

or #a^2+10ad+25d^2=a^2+13ad+22d^2#

i.e. #3d^2=3ad# or #d(d-a)=0# i.e. #a=d# as #d!=0#

As #4^(th)#, #8^(th)# and #16^(th)# terms are #a+3d#, #a+7d# and #a+15d#

We have #(a+3d)(a+15d)=4d xx 16d=64d^2=(8d)^2=(a+7d)^2#

Hence, #4^(th)#, #8^(th)# and #16^(th)# terms too are in GP.