JAMB 1990 · UME · Q28

Sequences & series (AP, GP)

Find the sum of the first 18 terms of the progression $3, 6, 12, \dots$

Worked solution (try it first)

This is a G.P. with $a = 3$ and $r = 6 \div 3 = 2$ .
Since $r > 1$ , use $S_n = \dfrac{a(r^n - 1)}{r - 1}$ with $n = 18$ .
So $S_{18} = \dfrac{3(2^{18} - 1)}{2 - 1}$
$= 3(2^{18} - 1)$ , option D.

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