WAEC 2022 · Paper 2 · Q9

Given that $\binom n4$ , $\binom n5$ and $\binom n6$ are the first 3 terms of a linear sequence (A.P.), find the:

Worked solution (try it first)

For an A.P., twice the middle term equals the sum of the other two: $2\binom n5 = \binom n4 + \binom n6$ .
Write each in terms of $\binom n4$ : $\binom n5 = \binom n4 \times \dfrac{n - 4}{5}$ and $\binom n6 = \binom n4 \times \dfrac{(n - 4)(n - 5)}{30}$ .
Divide by $\binom n4$ : $\dfrac{2(n - 4)}{5} = 1 + \dfrac{(n - 4)(n - 5)}{30}$ .
Multiply by 30: $12(n - 4) = 30 + (n - 4)(n - 5)$ , so $12n - 48 = n^2 - 9n + 50$ .
Rearrange: $n^2 - 21n + 98 = 0$ , so $(n - 7)(n - 14) = 0$ , and $n = 7$ or $n = 14$ .

$n = 14$ : the terms are $1001, 2002, 3003$ , so the common difference is $1001$ .
$n = 7$ : the terms are $35, 21, 7$ , so the common difference is $-14$ .