WAEC 2009 · Paper 2 · Q9

Sequences, series & binomial expansion

(a)
The 3rd and 6th terms of a geometric progression (G.P.) are 2 and 54 respectively. Find the: (i) common ratio; (ii) first term; (iii) sum of the first ten terms, correct to the nearest whole number.
Separate values with commas, e.g. 3, −2
(b)
The ratio of the coefficient of $x^4$ to that of $x^3$ in the binomial expansion of $(1 + 2x)^n$ is $3 : 1$ . Find the value of $n$ .

Worked solution (try it first)

(a)(i)

The $n$ th term of a G.P. is $ar^{n-1}$ , so $ar^2 = 2$ and $ar^5 = 54$ .
Divide the second equation by the first: $r^3 = 27$ , so $r = 3$ .

(ii)

Put $r = 3$ into $ar^2 = 2$ : $9a = 2$ , so $a = \frac29$ .

(iii)

$S_{10} = \dfrac{a(r^{10} - 1)}{r - 1}$
$= \dfrac{\frac29(3^{10} - 1)}{2}$ .
$3^{10} - 1 = 59\,048$ , so $S_{10} = \dfrac{59\,048}{9} = 6560.9$ , which is $6561$ to the nearest whole number.

(b)

The coefficient of $x^r$ in $(1 + 2x)^n$ is ${}^nC_r\,2^r$ .
Set up the ratio: $\dfrac{{}^nC_4\,2^4}{{}^nC_3\,2^3} = 3$ .
$\dfrac{{}^nC_4}{{}^nC_3} = \dfrac{n - 3}{4}$ , so the ratio is $2 \times \dfrac{n - 3}{4} = \dfrac{n - 3}{2}$ .
Solve $\dfrac{n - 3}{2} = 3$ : $n = 9$ .

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