JAMB 1987 · UME · Q29

Find the sum of the first 21 terms of the progression $-10, -8, -6, \dots$

Worked solution (try it first)

This is an A.P. with first term $a = -10$ and common difference $d = -8 - (-10) = 2$ .
Use $S_n = \frac n2\big(2a + (n - 1)d\big)$ with $n = 21$ : the bracket is $2(-10) + 20 \times 2$ , which is $-20 + 40 = 20$ .
So $S_{21} = \frac{21}{2} \times 20 = 210$ , option D.