2011 H2 Mathematics Paper 1 Question 9

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(i)

Depth drilled on 10th day =193 m{=193 \textrm{ m}}
Total depth =4810 m{=4810 \textrm{ m}}

(ii)

40 days{40 \textrm{ days}}

Full solutions

(i)

u10=a+(n−1)d=256+(10−1)(−7)=193\begin{align*} u_{10} &= a + \left( n - 1 \right) d \\ &= 256 + (10-1)(-7) \\ &= 193 \end{align*}
Depth drilled on 10th day =193 m  ■{= 193 \textrm{ m} \; \blacksquare}
un<10a+(n−1)d<10256+(n−1)(−7)<10\begin{align*} u_n &< 10 \\ a + \left( n - 1 \right) d &< 10 \\ 256 + (n-1)(-7) &< 10 \\ \end{align*}
7n>253n>36.14\begin{align*} 7n &> 253 \\ n &> 36.14 \end{align*}
Hence the last day of drilling is on the 37th day
S37=n2(2a+(n−1)d)=372(2(256)+36(−7))=4810\begin{align*} S_{37} &= \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) \\ &= \frac{37}{2} \Big( 2(256) + 36(-7) \Big) \\ &= 4810 \end{align*}
Total depth when drilling is completed =4810 m  ■{= 4810 \textrm{ m} \; \blacksquare}

(ii)

Theoretical maximum depth
S∞=a1−r=2561−89\begin{align*} S_{\infty} &= \frac{a}{1-r} \\ &= \frac{256}{1-\frac{8}{9}} \\ \end{align*}
Sn>0.99S∞256(1−(89)n)1−89>0.992561−891−(89)n>0.99(89)n<0.01nln⁡89<ln⁡0.01n>ln⁡0.01ln⁡89n>39.10\begin{align*} S_n &> 0.99 S_{\infty} \\ \frac{256\left(1-\left(\frac{8}{9}\right)^n\right)}{1-\frac{8}{9}} &> 0.99 \frac{256}{1-\frac{8}{9}} \\ 1-\left(\frac{8}{9}\right)^n &> 0.99 \\ \left(\frac{8}{9}\right)^n &< 0.01 \\ n \ln \frac{8}{9} &< \ln 0.01 \\ n &> \frac{\ln 0.01}{\ln \frac{8}{9}} \\ n &> 39.10 \end{align*}
Hence, to exceed 99% of the theoretical maximum depth,
it takes 40{40} days ■{\blacksquare}