WAEC 2010 · Paper 2 · Q13

(a)
The third term of a Geometric Progression (G.P.) is 24 and its seventh term is $4\frac{20}{27}$ . Find its first term.
(b)
Given that $y$ varies directly as $x$ and inversely as the square of $z$ . If $y = 4$ when $x = 3$ and $z = 1$ , find $y$ when $x = 3$ and $z = 2$ .

Worked solution (try it first)

(a)

The $n$ th term of a G.P. is $ar^{n-1}$ , so $ar^2 = 24$ and $ar^6 = 4\frac{20}{27} = \frac{128}{27}$ .
Divide the second equation by the first: $r^4 = \dfrac{128}{27} \div 24$
$= \dfrac{128}{648}$
$= \dfrac{16}{81}$ .
So $r^2 = \frac49$ (and $r = \pm\frac23$ ).
Then $a = \dfrac{24}{r^2}$
$= 24 \times \dfrac94$
$= 54$ .
The first term is 54.

(b)

Write the variation: $y = \dfrac{kx}{z^2}$ .
Put in $y = 4$ , $x = 3$ , $z = 1$ : $4 = 3k$ , so $k = \frac43$ .
Now $x = 3$ , $z = 2$ : $y = \dfrac{\frac43 \times 3}{2^2}$ , which is $\dfrac44 = 1$ .

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