2009 H2 Mathematics Paper 1 Question 8

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(ii)

L=272 cm{L = 272 \textrm{ cm}}
u13=10 cm{u_{13} = 10 \textrm{ cm}}

(iii)

d=−0.491{d=-0.491}
Length of longest bar =16.8 cm{=16.8 \textrm{ cm}}

Full solutions

(i)

u25=5arn−1=520r24=5r24=14r=(14)124=(12)112\begin{align*} u_{25} &= 5 \\ ar^{n-1} &= 5 \\ 20r^{24} &= 5 \\ r^{24} &= \frac{1}{4} \\ r &= \left(\frac{1}{4}\right)^\frac{1}{24} \\ &= \left(\frac{1}{2}\right)^\frac{1}{12} \\ \end{align*}
S∞=a1−r=201−(12)112=356.34<357\begin{align*} S_{\infty} &= \frac{a}{1-r} \\ &= \frac{20}{1-\left(\frac{1}{2}\right)^\frac{1}{12}} \\ &= 356.34 \\ &< 357 \end{align*}
Hence the total length of all the bars must be less than 357{357} cm, not matter how many bars there are ■{\blacksquare}

(ii)

L=S25=a(1−rn)1−r=20(1−(12)2512)1−(12)112=272 cm (3sf)  ■\begin{align*} L &= S_{25} \\ &= \frac{a(1-r^n)}{1-r} \\ &= \frac{20\Big(1-\left(\frac{1}{2}\right)^{\frac{25}{12}}\Big)}{1-\left(\frac{1}{2}\right)^{\frac{1}{12}}} \\ &= 272 \textrm{ cm (3sf)} \; \blacksquare \end{align*}
Length of 13{13}th bar
u13=ar12=20(12)=10 cm  ■\begin{align*} u_{13} &= ar^{12} \\ &= 20\left(\frac{1}{2}\right) \\ &= 10 \textrm{ cm} \; \blacksquare \end{align*}

(iii)

a+24d=5252(2a+24d)=272.26\begin{align} && \quad a + 24d &= 5 \\ && \quad \frac{25}{2}(2a+24d) &= 272.26 \\ \end{align}
Solving with a GC,
a=16.8 (3 sf)d=−0.491 (3 sf)  ■\begin{align*} a &= 16.8 \textrm{ (3 sf)} \\ d &= -0.491 \textrm{ (3 sf)} \; \blacksquare \end{align*}
Length of longest bar =16.8 cm  ■{=16.8 \textrm{ cm} \; \blacksquare}