2023 H2 Mathematics Paper 1 Question 11

Arithmetic and Geometric Progressions (APs, GPs)

Answers

Smallest ${{a= 513.89.}}$

{{400000(1.001)^n -}}

{{1000x (1.001^n - 1).}}

{{\$ 1323.63.}}

{{\$ 76508.52.}}

{{k = 287}}

{{y = 1321.54.}}

{{\$15990.}}

\begin{align*} S_{36} &= \frac{36}{2} \left( 2a + (36-1) 50 \right) \\ &= 18 ( 2 a + 1750 ) \end{align*}

\begin{gather*} 36 a + 31500 \geq 50000 \\ 36 a \geq 18500 \\ a \geq \frac{4625}{9} \end{gather*}

{ \text{Smallest } a = 513.89 \; \blacksquare }

Let ${{O_n}}$ denote the amount owed at the end of month ${{n}}$

\begin{align*} O_1 &= 400000 (1.001) - x \\ O_2 &= 1.001 (400000 (1.001) - x ) - x \\ &= 400000 (1.001)^2 - x(1.001) - x \\ O_3 &= 1.001 (400000 (1.001)^2 - x(1.001) - x) - x \\ &= 400000 (1.001)^3 - x(1.001)^2 - x(1.001) - x \\ & \cdots \\ O_n &= 400000 (1.001)^n - x(1.001)^{n-1} - \dotsb - x(1.001) - x \\ &= 400000(1.001)^n - \frac{x(1.001^n - 1)}{1.001-1} \\ &= 400000(1.001)^n - 1000x (1.001^n - 1) \; \blacksquare \end{align*}

(ci)

When ${{O_{360} = 0,}}$

\begin{gather*} 400000(1.001)^{360} - 1000x (1.001^{360} - 1) = 0 \end{gather*}

\begin{align*} x &= \frac{400000(1.001)^{360}}{1000 (1.001^{360} - 1)} \\ &= 1323.63 \; \blacksquare \end{align*}

(cii)

\begin{align*} & \text{Interest} \\ & = 360(1323.63) - 400000 \\ & = 76508.52 \; \blacksquare \end{align*}

(di)

When ${{x = 1600}}$ and ${{O_{n} = 0,}}$

\begin{gather*} 400000(1.001)^{n} - 1000(1600) (1.001^{n} - 1) = 0 \\ 400000(1.001)^{n} - 1600000(1.001)^n + 1600000 = 0 \\ -1200000 (1.001)^{n} = -1600000 \\ (1.001)^{n} = \frac{1600000}{1200000} \\ \end{gather*}

\begin{align*} n &= \frac{\ln \frac{4}{3}}{\ln 1.001} \\ &= 287.83 \\ k &= 287 \; \blacksquare \end{align*}

\begin{align*} & \text{Last payment} \\ & = O_{287} \times 1.001 \\ & = 1321.54 \; \blacksquare \end{align*}

(dii)

\begin{align*} & \text{Interest} \\ & = 287(1600.00) + \$1321.54 - 400000 \\ & = 60521.54 \end{align*}

\begin{align*} & \text{Savings} \\ &= 76508.52 - 60521.54 \\ &= \$15990 \; \blacksquare \end{align*}