WAEC 2020 · Paper 2 · Q3

Sequences, series & binomial expansion
Sequences & series (AP, GP)

The first and second terms of a linear sequence (A.P.) are $3$ and $8$ respectively. Determine the least number of terms of the A.P. that must be added so that the sum is greater than $250$ .

(a)
Least number of terms

Worked solution (try it first)

The first term is $a = 3$ and the common difference is $d = 8 - 3 = 5$ .
The sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$
$= \frac{n}{2}(5n + 1)$ .
The sum must be greater than 250: $\frac{n}{2}(5n + 1) > 250$ , so $5n^2 + n - 500 > 0$ .
Solve $5n^2 + n - 500 = 0$ with the formula: $n = \frac{-1 \pm \sqrt{1 + 10\,000}}{10}$ , so $n \approx 9.90$ (the negative root doesn't apply to a number of terms).
$n$ must be a whole number greater than 9.90, so the least number of terms is $10$ .
Check: $S_9 = \frac92 \times 46 = 207$ , which is not enough, and $S_{10} = 5 \times 51 = 255$ , which is greater than 250 ✓.

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