2024 H2 Mathematics Paper 1 Question 5

Arithmetic and Geometric Progressions (APs, GPs)

Answers

$27 r^{3} - 27 r^{2} + 4 = 0.$

$r = - \frac{1}{3},$ $a = 24.$

$k = 4.$

(a)

\begin{array}{lcr} S_{\infty} = 18 \\ \frac{a}{1 - r} = 18 \\ a = 18 - 18 r \end{array}

\begin{array}{lcr} u_{3} = \frac{8}{3} \\ a r^{2} = \frac{8}{3} \\ 3 a r^{2} = 8 \end{array}

Substituting $(3)$ into $(6),$

\begin{matrix} 3 (18 - 18 r) r^{2} = 8 \\ 54 r^{2} - 54 r^{3} = 8 \\ 27 r^{2} - 27 r^{3} = 4 \\ 27 r^{3} - 27 r^{2} + 4 = 0 ∎ \end{matrix}

(b)

From GC,

r = \frac{2}{3} or r = - \frac{1}{3}

Since $r < 0,$ $r = - \frac{1}{3} ∎$

Substituting $r = - \frac{1}{3}$ into $(3),$ $a = 24 ∎$

(c)

\begin{matrix} 80 \sum_{r = k + 1}^{\infty} u_{r} = \sum_{r = 1}^{k} u_{r} \\ 80 (\sum_{r = 1}^{\infty} u_{r} - \sum_{r = 1}^{k} u_{r}) = \sum_{r = 1}^{k} u_{r} \\ 80 \sum_{r = 1}^{\infty} u_{r} - 80 \sum_{r = 1}^{k} u_{r} = \sum_{r = 1}^{k} u_{r} \\ 80 \sum_{r = 1}^{\infty} u_{r} = 81 \sum_{r = 1}^{k} u_{r} \\ \frac{80 a}{1 - r} = \frac{81 a (1 - r^{k})}{1 - r} \\ 80 = 81 (1 - r^{k}) \\ 80 = 81 - 81 r^{k} \\ 81 r^{k} = 1 \\ r^{k} = \frac{1}{81} \\ {(- \frac{1}{3})}^{k} = \frac{1}{81} \\ {(- \frac{1}{3})}^{k} = {(- \frac{1}{3})}^{4} \\ k = 4 ∎ \end{matrix}