WAEC Further Mathematics Past Questions & Answers - Page 125

$\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27$

$(m^{2} - 4m + 4) - (m^{2} + 5m + 4) = -27$

$-9m = -27 \implies m = 3$

$T_{n} = a + (n - 1)d$ (for a linear or arithmetic progression)

Given: $T_{n} = 71, a = -6, d = -2\frac{1}{2} - (-6) = 3\frac{1}{2}$

$\implies 71 = -6 + (n - 1)\times 3\frac{1}{2}$

$71 = -6 + 3\frac{1}{2}n - 3\frac{1}{2} = -9\frac{1}{2} + 3\frac{1}{2}n$

$71 + 9\frac{1}{2} = 3\frac{1}{2}n  \implies  n = \frac{80\frac{1}{2}}{3\frac{1}{2}}$

$= 23$

$T_{n} = ar^{n-1}$ (for a geometric progression)

$T_{3} = ar^{3-1} = ar^{2} = \frac{8}{3}$

$T_{6} = ar^{6-1} = ar^{5} = \frac{64}{81}$

Dividing $T_{6}$ by $T_{3}$,

$\frac{ar^{5}}{ar^{2}} = \frac{\frac{64}{81}}{\frac{8}{3}} \implies r^{3} = \frac{8}{27}$

$\therefore r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$

Listing out in order, the terms of this expansion have coefficients: $^{6}C_{6}, ^{6}C_{5}, ^{6}C_{4}, ^{6}C_{3}, ^{6}C_{2}, ^{6}C_{1}, ^{6}C_{0}$

The 4th term = $^{6}C_{3}(3x)^{3}(-y)^{3}  = 20 \times 27x^{3} \times -y^{3}$

= $-540x^{3}y^{3}$.

$[\frac{1}{3}(2 + x)]^{6} = (\frac{2}{3} + \frac{x}{3})^{6}$

The coefficient of $x^{3}$ is 

$^{6}C_{3}(\frac{2}{3})^{3}(\frac{1}{3})^{3}x^{3} = (\frac{6!}{3!3!})(\frac{8}{27})(\frac{1}{27})x^{3}$

= $\frac{160}{729}$