WAEC 2021 · Paper 2 · Q7

(a)
A three-digit number is such that the sum of the digits is 12 and the last digit is three times the first one. If the middle digit is half the sum of the other two, find the number.
(b)
The 7th and 18th terms of an Arithmetic Progression (A.P.) are $15\frac14$ and $42\frac34$ respectively. Find the: (i) $n$ th term; (ii) sixth term, of the progression.

Worked solution (try it first)

(a)

Let the digits be $x$ (first), $y$ (middle) and $z$ (last).
The last digit is three times the first: $z = 3x$ .
The middle digit is half the sum of the other two: $y = \frac{x + 3x}{2} = 2x$ .
The digits add up to 12: $x + 2x + 3x = 12$ , so $6x = 12$ and $x = 2$ .
Then $y = 4$ and $z = 6$ .
The number is 246.

(b)(i)

The $n$ th term is $U_n = a + (n - 1)d$ .
So $a + 6d = 15\frac14 = \frac{61}{4}$ and $a + 17d = 42\frac34 = \frac{171}{4}$ .
Take the first from the second: $11d = \frac{110}{4}$ , so $d = \frac{10}{4} = \frac52$ .
Then $a = \frac{61}{4} - 6 \times \frac52$
$= \frac{61}{4} - 15$
$= \frac14$ .
So $U_n = \frac14 + \frac52(n - 1)$
$= \frac{1 + 10n - 10}{4}$
$= \frac{10n - 9}{4}$ .

(ii)

$U_6 = \frac{60 - 9}{4}$
$= \frac{51}{4}$
$= 12\frac34$ .

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