WAEC 2024 · Paper 2 · Q6

Given that $(x + 2)$ , $(4x + 3)$ and $(7x + 24)$ are consecutive terms of a geometric progression (G.P.), find the:

(a)
values of $x$ ;
Separate values with commas, e.g. 3, −2
(b)
common ratio (for each value of $x$ ).
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

For consecutive terms of a G.P., the ratios are equal: $\frac{4x + 3}{x + 2} = \frac{7x + 24}{4x + 3}$ .
Cross-multiply: $(4x + 3)^2 = (x + 2)(7x + 24)$ .
Expand both sides: $16x^2 + 24x + 9 = 7x^2 + 38x + 48$ .
Collect everything on one side: $9x^2 - 14x - 39 = 0$ .
Factorise: $(x - 3)(9x + 13) = 0$ , so $x = 3$ or $x = -\frac{13}{9}$ .

(b)

When $x = 3$ the terms are $5, 15, 45$ , so the common ratio is $\frac{15}{5} = 3$ .
When $x = -\frac{13}{9}$ : $x + 2 = \frac59$ , $4x + 3 = -\frac{52}{9} + \frac{27}{9}$
$= -\frac{25}{9}$ and $7x + 24 = -\frac{91}{9} + \frac{216}{9}$
$= \frac{125}{9}$ .
The common ratio is $-\frac{25}{9} \div \frac59 = -5$ .

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