WAEC 2018 · Paper 2 · Q1

1. Write each fact as an equation.
2. The sum of the first ten terms: $S_{10} = \frac{10}{2}[2a + 9d]$
   $= 5(2a + 9d)$
   $= 130$, so $2a + 9d = 26$.
3. The fifth term is 3 times the first: $a + 4d = 3a$, so $4d = 2a$ and $a = 2d$.

(a)

1. Put $a = 2d$ into $2a + 9d = 26$: $4d + 9d = 26$, so $13d = 26$ and the common difference is $d = 2$.

(b)

1. The first term is $a = 2d = 4$.
2. (Check: $5(8 + 18) = 130$ ✓ and $T_5 = 4 + 8 = 12 = 3 \times 4$ ✓.)

(c)

1. The last term is 28: $a + (n - 1)d = 28$, so $4 + 2(n - 1) = 28$, $2(n - 1) = 24$, $n - 1 = 12$ and there are $n = 13$ terms.