2018 H2 Mathematics Paper 1 Question 11

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(ia)
$102.43{\$102.43}
(ib)
$1215.71{\$1215.71}
(ic)
n=30{n=30}
First day of June 2018
(iia)
$(100+12b){\$(100+12b)}
(iib)
b=43{b=\frac{4}{3}}

(iii)

b=1.23{b=1.23}

Full solutions

(ia)
Amount worth at end of 31 December 2016
u12=arn−1=(100(1.002))(1.002)11=102.43  ■\begin{align*} u_{12} &= ar^{n-1} \\ &= \Big( 100(1.002) \Big) (1.002)^{11} \\ &= 102.43 \; \blacksquare \end{align*}
(ib)
Amount worth at end of 31 December 2016
S12=a(rn−1)r−1=(100(1.002))((1.002)12−1)1.002−1=1215.71  ■\begin{align*} S_{12} &= \frac{a\left(r^{n}-1\right)}{r-1} \\ &= \frac{\Big( 100(1.002) \Big) \Big( (1.002)^{12}-1 \Big) }{1.002-1} \\ &= 1215.71 \; \blacksquare \end{align*}
(ic)
Sn>3000a(rn−1)r−1>3000(100(1.002))((1.002)n−1)1.002−1>3000(1.002)n−1>0.05988nln⁡1.002>ln⁡1.0599n>ln⁡1.0599ln⁡1.002n>29.107\begin{gather*} S_n > 3000 \\ \frac{a\left(r^{n}-1\right)}{r-1} > 3000 \\ \frac{\Big( 100(1.002) \Big)\Big( \left(1.002\right)^n - 1 \Big)}{1.002-1} > 3000 \\ \left(1.002\right)^n - 1 > 0.05988 \\ n \ln 1.002 > \ln 1.0599 \\ n > \frac{\ln 1.0599}{\ln 1.002} \\ n > 29.107 \end{gather*}
At n=30,{n=30,} (last day of June 2018),
S30=3094.82S_{30} = 3094.82
At n=29,{n=29,} (last day of May 2018),
S29=2988.65S_{29} = 2988.65
On first day of June 2018, he will have $3088.65{\$3088.65}
Hence the total in the account will first exceed $3000{\$3000} on the first day of June 2018 ■{\blacksquare}
(iia)
Amount worth at end of 31 December 2016
=$(100+12b)  ■= \$(100+12b) \; \blacksquare
(iib)
Amount worth at end of 31 December 2017
=2400+b+2b+3b+…+24b=2400+242(2b+(24−1)b)=2400+300b\begin{align*} &= 2400 + b + 2b + 3b + \ldots + 24b \\ &= 2400 + \frac{24}{2}\Big( 2b + (24-1)b \Big) \\ &= 2400 + 300b \end{align*}
2400+300b=2800b=43  ■\begin{align*} 2400 + 300 b &= 2800 \\ b = \frac{4}{3} \; \blacksquare \end{align*}

(iii)

6000+n2(2a+(n−1)d)=a′(rn−1)r−16000 + \frac{n}{2} \Big(2a + (n-1)d \Big) = \frac{a'(r^n-1)}{r-1}
6000+602(2b+(60−1)d)=(100(1.01))(1.0160−1)1.01−16000 + \frac{60}{2} \Big(2b + (60-1)d \Big) = \frac{\Big(100(1.01)\Big)\Big( 1.01^{60} - 1 \Big)}{1.01-1}
1830b=2248.6b=1.23  ■\begin{align*} 1830b &= 2248.6 \\ b &= 1.23 \; \blacksquare \end{align*}