WAEC 2019 · Paper 2 · Q2

1. The $n$th term of an A.P. is $a + (n - 1)d$.
2. So $T_2 = a + d = x - 1$, $T_4 = a + 3d = x + 1$ and $T_6 = a + 5d = 7$.

(a)

1. Take the second term from the fourth: $(a + 3d) - (a + d) = (x + 1) - (x - 1)$.
2. Both $a$ and $x$ cancel: $2d = 2$, so the common difference is $d = 1$.

(b)

1. Put $d = 1$ into $a + 5d = 7$: $a + 5 = 7$, so the first term is $a = 2$.

(c)

1. Put $a = 2$ and $d = 1$ into $a + d = x - 1$: $3 = x - 1$, so $x = 4$.
2. (Check: $T_4 = a + 3d = 5$ and $x + 1 = 5$ ✓.)