WAEC 2008 · Paper 2 · Q9

(a)
The sum of the first $n$ terms of a sequence is given by $S_n = \dfrac{5n^2}{2} + \dfrac{5n}{2}$ . Write down the first four terms of the sequence and find an expression for the $n$ th term.
(b)(i)
The equation of a circle is given by $x^2 + y^2 - 10x - 8y + 25 = 0$ . Show that the circle touches the $x$ -axis.
Model answer
Complete the squares: $(x - 5)^2 + (y - 4)^2 = 16$ , so the centre is $(5, 4)$ and the radius is 4. The centre is 4 units above the $x$ -axis, which equals the radius, so the circle touches the $x$ -axis. (Or: put $y = 0$ to get $x^2 - 10x + 25 = 0$ , which is $(x - 5)^2 = 0$ , a repeated root.)
(b)(ii)
Find the coordinates of the point of contact.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$S_1 = \frac52 + \frac52 = 5$ , so $T_1 = 5$ .
$S_2 = 10 + 5 = 15$ , so $T_2 = S_2 - S_1 = 10$ .
$S_3 = 22.5 + 7.5 = 30$ , so $T_3 = 30 - 15 = 15$ .
$S_4 = 40 + 10 = 50$ , so $T_4 = 50 - 30 = 20$ .
The first four terms are $5, 10, 15, 20$ : an AP with $a = 5$ and $d = 5$ .
$n$ th term: $T_n = a + (n - 1)d = 5 + 5(n - 1)$ , which simplifies to $T_n = 5n$ .

Put $y = 0$ (the $x$ -axis): $x^2 - 10x + 25 = 0$ .
This is $(x - 5)^2 = 0$ , a repeated root, so the circle meets the $x$ -axis at one point only: it touches it.