2012 H2 Mathematics Paper 2 Question 4

Arithmetic and Geometric Progressions (APs, GPs)

Answers

{24}

months. 1 December 2002

{20100(1.005^n-1)}

{45}

months. September 2004

{1.80\%}

\begin{gather*} S_n > 5000 \\ \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) > 5000 \\ \frac{n}{2}\Big(2(100) + (n-1)(10)\Big) > 5000 \\ 5 n^2 + 95 n - 5000 > 0 \\ \end{gather*}

{n > 23.52} \; \allowbreak \textrm{ or } \; \allowbreak {n < -42.52 \textrm{ (NA)}}

Hence least

{n=24}

Account first becomes greater than $5000 on 1 December 2002

{\blacksquare}

Considering the total amount in his account:

Start of

{1}

st month:

100

End of

{1}

st month:

100(1.005)

Start of

{2}

nd month:

100(1.005)+100

End of

{2}

nd month:

100(1.005)^2+100(1.005)

Start of

{3}

rd month:

100(1.005)^2+100(1.005)+100

End of

{3}

rd month:

100(1.005)^3+100(1.005)^2+100(1.005)

{\quad \cdots}

End of

{n}

th month:

100(1.005)^n+100(1.005)^{n-1}+\ldots+100(1.005)

This form a geometric series with first term

{a=100(1.005)}

and common ratio

{r=1.005}

Value of account on the last day of

{n\textrm{th}}

month

\begin{align*} S_n &= \frac{a\left(1-r^{n}\right)}{1-r} \\ &= \frac{100(1.005)(1-1.005^n)}{1-1.005} \\ &= 20100(1.005^n-1) \; \blacksquare \end{align*}

When it is greater than $5000,

20100(1.005^n-1) > 5000

\begin{align*} 1.005^n - 1 &> \frac{50}{201} \\ 1.005^n &> \frac{251}{201} \\ n &> \frac{\ln \frac{251}{201}}{\ln 1.005} \\ n &> 44.54 \end{align*}

Least

{n=45}

Account will first be greater than $5000 in September 2004

{\blacksquare}

We will replace

{1.005}

in the working in (ii) with

{x}

For the account to be $5000 on 2 December 2003, it will be

{5000x}

on 31st December 2003 (end of 36 months)

\begin{align*} S_n &= 5000x \\ \frac{a\left(1-r^{n}\right)}{1-r} &= 5000x \\ \frac{100x(1-x^{36})}{1-x} &= 5000x \\ \end{align*}

Solving with a GC,

{x=1.01796}

Hence the interest rate must be

{1.80\% \; \blacksquare}