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

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Answer 1 · 2017-12-04T11:08:46

Please refer to the Discussion in The 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.)#