WAEC 2017 · Paper 2 · Q2✱

(a)
The $n$ th term of an Arithmetic Progression (A.P.) is given by $U_n = (3n - 2)\log5$ . Find the sum of the first $n$ terms of the Arithmetic Progression in terms of $\log5$ .

Worked solution (try it first)

Find the first two terms: $U_1 = (3 - 2)\log 5 = \log 5$ and $U_2 = (6 - 2)\log 5 = 4\log 5$ .
The first term is $a = \log 5$ and the common difference is $d = 4\log 5 - \log 5 = 3\log 5$ .
Use $S_n = \frac n2[2a + (n - 1)d]$ : $S_n = \frac n2[2\log 5 + 3(n - 1)\log 5]$ .
Simplify the bracket: $2 + 3n - 3 = 3n - 1$ , so $S_n = \dfrac{n(3n - 1)}{2}\log 5$ .