JAMB 2011 · UTME · Q22

The second term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms.

Worked solution (try it first)

The 2nd term is $ar = 4$ and the 4th is $ar^3 = 16$ .
Divide: $r^2 = 4$ , so take $r = 2$ and then $a = 2$ .
Use $S_n = \dfrac{a(r^n - 1)}{r - 1}$ : $S_5 = \dfrac{2(2^5 - 1)}{2 - 1}$ , which is $2 \times 31 = 62$ .
So the sum is 62, option B.
(With $r = -2$ the sum is $-22$ , which is not an option.)