NECO 2024 · Paper 2 · Q9

(a)
The 9th term of an arithmetic progression is 61 while the 21st term is 145. Find the sum of the first 48 terms.
(b)
Evaluate $\displaystyle\int_0^2 \left(\tfrac32x^2 + 5\right)dx$ .

Worked solution (try it first)

(a)

The $n$ th term of an A.P. is $a + (n - 1)d$ .
So $T_9 = a + 8d = 61$ and $T_{21} = a + 20d = 145$ .
Take the first equation from the second: $12d = 84$ , so $d = 7$ .
Then $a = 61 - 8 \times 7 = 61 - 56 = 5$ .
$S_n = \frac{n}{2}[2a + (n - 1)d]$ , so $S_{48} = \frac{48}{2}[2 \times 5 + 47 \times 7]$
$= 24 \times (10 + 329)$
$= 24 \times 339$
$= 8136$ .

(b)

Integrate term by term: $\frac32x^2$ becomes $\frac32 \times \frac{x^3}{3} = \frac{x^3}{2}$ , and $5$ becomes $5x$ .
So $\displaystyle\int_0^2 \left(\tfrac32x^2 + 5\right)dx = \left[\frac{x^3}{2} + 5x\right]_0^2$ .
Put in the limits: $\left(\frac{8}{2} + 10\right) - (0 + 0) = 4 + 10$
$= 14$ .

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