JAMB 1998 · UME · Q19

If $p + 1$ , $2p - 10$ , $1 - 4p^2$ are consecutive terms of an arithmetic progression, find the possible values of $p$ .

Worked solution (try it first)

In an A.P. the middle term is the average of its neighbours, so $2(2p - 10) = (p + 1) + (1 - 4p^2)$ .
Tidy up: $4p - 20 = 2 + p - 4p^2$ , so $4p^2 + 3p - 22 = 0$ .
Factorise: $(4p + 11)(p - 2) = 0$ .
So $p = -\frac{11}{4}$ or $p = 2$ , option C.