2008 H2 Mathematics Paper 1 Question 10

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(i)

n=34{n=34}
1 October 2011
(iia)
$6.08{\$6.08}
(iib)
$310.30{\$310.30}
(iic)
81{81} complete months

Full solutions

(i)

Sn>2000n2(2a+(n−1)d)>2000n2(2(10)+(n−1)(3))>200032n2+172n−2000>0n>33.79 or n<−39.46 (NA)\begin{gather*} S_n > 2000 \\ \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) > 2000 \\ \frac{n}{2} \Big( 2(10) + \left( n - 1 \right) \left(3\right) \Big) > 2000 \\ \frac{3}{2} n^2 + \frac{17}{2} n - 2000 > 0 \\ n> 33.79 \textrm{ or } n< -39.46 \textrm{ (NA)} \end{gather*}
Least n=34{n=34}
She will have first saved over $2000 on 1 October 2011 ■{\blacksquare}
(iia)
Total amount due to original $10 at the end of 2 years
u24=arn−1=(10(1.02))(1.02)23=16.08\begin{align*} u_{24} &= ar^{n-1} \\ &= \Big( 10(1.02) \Big) (1.02)^{23} \\ &= 16.08 \end{align*}
Compound interest due to original $10 at the end of 2 years
=16.08−10=$6.08  ■=16.08-10 = \$6.08 \;\blacksquare
(iib)
Total amount at end of 2 years
S24=a(rn−1)r−1=(10(1.02))((1.02)24−1)1.02−1=$310.30  ■\begin{align*} S_{24} &= \frac{a\left(r^{n}-1\right)}{r-1} \\ &= \frac{\Big( 10(1.02) \Big) \Big( (1.02)^{24}-1 \Big) }{1.02-1} \\ &= \$310.30 \; \blacksquare \end{align*}
(iic)
Sn>2000a(rn−1)r−1>2000(10(1.02))((1.02)n−1)1.02−1>2000(1.02)n−1>3.9216nln⁡1.02>ln⁡4.9216n>ln⁡4.9216ln⁡1.02n>80.476\begin{gather*} S_n > 2000 \\ \frac{a\left(r^{n}-1\right)}{r-1} > 2000 \\ \frac{\Big( 10(1.02) \Big)\Big( \left(1.02\right)^n - 1 \Big)}{1.02-1} > 2000 \\ \left(1.02\right)^n - 1 > 3.9216 \\ n \ln 1.02 > \ln 4.9216 \\ n > \frac{\ln 4.9216}{\ln 1.02} \\ n > 80.476 \end{gather*}
Hence the total in the account first exceed $2000 after 81{81} complete months ■{\blacksquare}