Specimen 2017 H2 Mathematics Paper 1 Question 4

Arithmetic and Geometric Progressions (APs, GPs)

Answers

(a)

1277 cm.{1277 \textrm{ cm}.}

(b)

85{85} bounces.
3400 cm.{3400 \textrm{ cm}.}

Full solutions

(a)

The downward movements form a GP with first term 200{200} and common ratio 89{\frac{8}{9}} while the upward movements form a GP with first term 16009{\frac{1600}{9}} and common ratio 89.{\frac{8}{9}.}
Total distance=S4,downwards+S4,upwards=200(1−(89)4)1−89+16009(1−(89)4)1−89=83810006561=1277 cm (nearest cm)  ■\begin{align*} & \textrm{Total distance} \\ & = S_{4,\textrm{downwards}} + S_{4,\textrm{upwards}} \\ & = \frac{200\Big( 1-\left(\frac{8}{9}\right)^{4} \Big)}{1-\frac{8}{9}} + \frac{\frac{1600}{9}\Big( 1-\left(\frac{8}{9}\right)^{4} \Big)}{1-\frac{8}{9}} \\ & = \frac{8381000}{6561} \\ & = 1277 \textrm{ cm (nearest cm)} \; \blacksquare \end{align*}
Total distance=S85,downwards+S84,upwards=200(1−(89)85)1−89+16009(1−(89)84)1−89=3399.84=3400 cm (nearest cm)  ■\begin{align*} & \textrm{Total distance} \\ & = S_{85,\textrm{downwards}} + S_{84,\textrm{upwards}} \\ & = \frac{200\Big( 1-\left(\frac{8}{9}\right)^{85} \Big)}{1-\frac{8}{9}} + \frac{\frac{1600}{9}\Big( 1-\left(\frac{8}{9}\right)^{84} \Big)}{1-\frac{8}{9}} \\ & = 3399.84 \\ & = 3400 \textrm{ cm (nearest cm)} \; \blacksquare \end{align*}

(b)

When the ball have stopped bouncing,
un<0.01200(89)n<0.01(89)n<120000ln⁡(89)n<ln⁡120000n>ln⁡120000ln⁡89n>84.082\begin{align*} u_n &< 0.01 \\ 200 \left( \frac{8}{9} \right)^n &< 0.01 \\ \left( \frac{8}{9} \right)^n &< \frac{1}{20000} \\ \ln \left( \frac{8}{9} \right)^n &< \ln \frac{1}{20000} \\ n &> \frac{\ln \frac{1}{20000}}{\ln \frac{8}{9}} \\ n &> 84.082 \\ \end{align*}
Hence the ball has made 85{85} bounces. ■{\blacksquare}