2023 H2 Mathematics Paper 1 Question 11

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(a)

Smallest a=513.89.a= 513.89.

(b)

400000(1.001)n−400000(1.001)^n - 1000x(1.001n−1).1000x (1.001^n - 1).
(ci)
$1323.63.\$ 1323.63.
(cii)
$76508.52.\$ 76508.52.
(di)
k=287k = 287
y=1321.54.y = 1321.54.
(dii)
$15990.\$15990.

Full solutions

(a)

S36=362(2a+(36−1)50)=18(2a+1750)\begin{align*} S_{36} &= \frac{36}{2} \left( 2a + (36-1) 50 \right) \\ &= 18 ( 2 a + 1750 ) \end{align*}
36a+31500≥5000036a≥18500a≥46259\begin{gather*} 36 a + 31500 \geq 50000 \\ 36 a \geq 18500 \\ a \geq \frac{4625}{9} \end{gather*}
Smallest a=513.89  ■{ \text{Smallest } a = 513.89 \; \blacksquare }

(b)

Let OnO_n denote the amount owed at the end of month nn

O1=400000(1.001)−xO2=1.001(400000(1.001)−x)−x=400000(1.001)2−x(1.001)−xO3=1.001(400000(1.001)2−x(1.001)−x)−x=400000(1.001)3−x(1.001)2−x(1.001)−x⋯On=400000(1.001)n−x(1.001)n−1−⋯−x(1.001)−x=400000(1.001)n−x(1.001n−1)1.001−1=400000(1.001)n−1000x(1.001n−1)  ■\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 O360=0,O_{360} = 0,

400000(1.001)360−1000x(1.001360−1)=0\begin{gather*} 400000(1.001)^{360} - 1000x (1.001^{360} - 1) = 0 \end{gather*}
x=400000(1.001)3601000(1.001360−1)=1323.63  ■\begin{align*} x &= \frac{400000(1.001)^{360}}{1000 (1.001^{360} - 1)} \\ &= 1323.63 \; \blacksquare \end{align*}

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

(di)

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

400000(1.001)n−1000(1600)(1.001n−1)=0400000(1.001)n−1600000(1.001)n+1600000=0−1200000(1.001)n=−1600000(1.001)n=16000001200000\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*}
n=ln⁡43ln⁡1.001=287.83k=287  ■\begin{align*} n &= \frac{\ln \frac{4}{3}}{\ln 1.001} \\ &= 287.83 \\ k &= 287 \; \blacksquare \end{align*}
Last payment=O287×1.001=1321.54  ■\begin{align*} & \text{Last payment} \\ & = O_{287} \times 1.001 \\ & = 1321.54 \; \blacksquare \end{align*}

(dii)
Interest=287(1600.00)+$1321.54−400000=60521.54\begin{align*} & \text{Interest} \\ & = 287(1600.00) + \$1321.54 - 400000 \\ & = 60521.54 \end{align*}
Savings=76508.52−60521.54=$15990  ■\begin{align*} & \text{Savings} \\ &= 76508.52 - 60521.54 \\ &= \$15990 \; \blacksquare \end{align*}