If the second, fourth and ninth term of an arithmetic progression are in geometric progression, then what is the common ratio of the GP?

1 Answer
Dec 4, 2017

Please refer to the Discussion in The Explanation.

Explanation:

Let the AP be, a, a+d, a+2d,...,a+(n-1)d,... (n in NN.).

Then, by what is given, a+d, (a+3d) and (a+8d) are in GP.

:. (a+3d)^2=(a+d)(a+8d).

:. a^2+6ad+9d^2=a^2+9ad+8d^2, or,

d^2-3ad=0, i.e.,

d(d-3a)=0.

:. d=0, or, d=3a.

If d=0, then, the GP becomes the constant sequence

a,a,a,...a,... for which the common ratio is 1. (a ne 0)

In case, d=3a, then the GP is 4a,10a,25a for which the

common ratio is 2.5a, or, 5/2a, (a ne 0.)