JAMB 1990 · UME · Q27

Find the sum of the first twenty terms of the arithmetic progression $\log a, \log a^2, \log a^3, \dots$

Worked solution (try it first)

By the laws of logs, $\log a^k = k\log a$ , so the terms are $\log a, 2\log a, 3\log a, \dots$
The sum of the first twenty is $(1 + 2 + \dots + 20)\log a$ , and $1 + 2 + \dots + 20 = \frac{20 \times 21}{2}$
$= 210$ .
So the sum is $210\log a = \log a^{210}$ , option D.