2013 H2 Mathematics Paper 1 Question 7

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(i)

A=1,  B=6,  C=−1,  D=1{A=1,\;}\allowbreak {B=6,\;}\allowbreak {C=-1,\;}\allowbreak {D=1}

(iii)

12{12} pieces

Full solutions

(i)

un=p128(23)n−1=p\begin{gather*} u_n = p \\ 128 \left( \frac{2}{3} \right)^{n - 1} = p \\ \end{gather*}
ln⁡p=ln⁡(128(23)n−1)=ln⁡128+ln⁡(23)n−1=ln⁡27+(n−1)(ln⁡2−ln⁡3)=(n+6)ln⁡2+(−n+1)ln⁡3  ■\begin{align*} \ln p &= \ln \left(128 \left( \frac{2}{3} \right)^{n - 1}\right) \\ &= \ln 128 + \ln \left(\frac{2}{3}\right)^{n-1} \\ &= \ln 2^7 + (n-1) \left( \ln 2 - \ln 3 \right) \\ &= (n+6)\ln 2 + (-n+1)\ln 3 \; \blacksquare \end{align*}
A=1,  B=6,  C=−1,  D=1  ■{A=1,\;}\allowbreak {B=6,\;}\allowbreak {C=-1,\;}\allowbreak {D=1} \;\blacksquare

(ii)

Theoretical maximum total length
S∞=a1−r=1281−23=384\begin{align*} S_\infty &= \frac{a}{1-r} \\ &= \frac{128}{1-\frac{2}{3}} \\ &= 384 \end{align*}
Hence the total length of string cut off can never be greater than 384 cm  ■{384 \textrm{ cm} \; \blacksquare}

(iii)

Sn>380a(1−rn)1−r>380128(1−(23)n)1−23>3801−(23)n>9596(23)n<196ln⁡(23)n<ln⁡196n>ln⁡196ln⁡23n>11.26\begin{align*} S_n &> 380 \\ \frac{a\left(1-r^{n}\right)}{1-r} &> 380 \\ \frac{128\left(1-\left(\frac{2}{3}\right)^n\right)}{1-\frac{2}{3}} &> 380 \\ 1-\left(\frac{2}{3}\right)^n &> \frac{95}{96} \\ \left(\frac{2}{3}\right)^n &< \frac{1}{96} \\ \ln \left(\frac{2}{3}\right)^n &< \ln \frac{1}{96} \\ n &> \frac{\ln \frac{1}{96}}{\ln \frac{2}{3}} \\ n &> 11.26 \\ \end{align*}

Hence the number of pieces that must be cut is 12  ■{12 \; \blacksquare}