JAMB 1993 · UME · Q23

If $k + 1$ , $2k - 1$ , $3k + 1$ are three consecutive terms of a geometric progression, find the possible values of the common ratio.

Worked solution (try it first)

For three terms of a G.P., the middle term squared equals the product of the outer two: $(2k - 1)^2 = (k + 1)(3k + 1)$ .
Expand: $4k^2 - 4k + 1 = 3k^2 + 4k + 1$ , so $k^2 - 8k = 0$ and $k(k - 8) = 0$ .
If $k = 0$ the terms are $1, -1, 1$ with ratio $-1$ .
If $k = 8$ they are $9, 15, 25$ with ratio $\frac{15}{9} = \frac53$ .
So the common ratio is $-1$ or $\frac53$ , option B.