2014 H2 Mathematics Paper 2 Question 3

Arithmetic and Geometric Progressions (APs, GPs)

Answers

{440 \textrm{ m}}

{(4 n^2 + 4 n) \textrm{ m}}

{35}

stages

{8 \left( 2^n - 1 \right) \textrm{ m}}

Travelling away from

{O}

1816 m from

{O}

(ia)

The distances form an AP with first term

{8}

and common difference

{8}

\begin{align*} S_{10} &= \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) \\ &= \frac{10}{2} \Big( 2(8) + (10-1)(8) \Big) \\ &= 440 \textrm{ m} \; \blacksquare \end{align*}

(ib)

Distance run

\begin{align*} S_{n} &= \frac{n}{2} \Big( 2a + \left( n - 1 \right) d \Big) \\ &= \frac{n}{2} \Big( 2(8) + (n-1)(8) \Big) \\ &= 4 n^2 + 4 n \; \blacksquare \end{align*}

When the athlete runs at least

{5}

km,

\begin{gather*} 4 n^2 + 4 n \geq 5000 \\ 4 n^2 + 4 n - 5000 \geq 0 \\ n\geq 34.86 \; \textrm{ or } \; n\leq -35.86 \textrm{ (NA)} \end{gather*}

Least number of stages

{= 35 \; \blacksquare}

The distances form an GP with first term

{8}

and common ratio

{2}

\begin{align*} S_{n} &= \frac{a\left(r^{n}-1\right)}{r-1} \\ &= \frac{8\Big( \left(2\right)^{n} - 1 \Big)}{2-1} \\ &= 8 \left( 2^n - 1 \right) \textrm{ m} \; \blacksquare \end{align*}

When the athlete runs exactly

{10}

km,

\begin{align*} 8 \left( 2^n - 1 \right) &= 10000 \\ 2^n &= 1251 \\ n &= \frac{\ln 1251}{\ln 2} \\ n &= 10.29 \end{align*}

Hence he has completed

{10}

complete stages

\begin{align*} S_{10} &= \frac{a\left(r^{n}-1\right)}{r-1} \\ &= \frac{8\Big( \left(2\right)^{10} - 1 \Big)}{2-1} \\ &= 8184 \end{align*}

He will run

{10,000-8184=1816}

meters in stage

{11}

\begin{align*} u_{11} &= ar^{n-1} \\ &= 8 \left( 2 \right)^{11-1} \\ &= 8192 \end{align*}

Since

{1816}

is less than half of

{u_{11},}

he is travelling away from

{O \; \blacksquare}

Distance from

{O: 1816 \textrm{ m} \; \blacksquare}