2016 H2 Maths Paper 1 Q4 — Worked Solution

Part (i)

4 marks

Approach

Four unknowns $a,d,b,r$ appear but only their relationship matters. Writing the three given equalities and subtracting consecutive pairs eliminates $a$ , leaving two equations in $d$ alone; dividing those two equations eliminates $d$ as well, leaving a single equation in $r$ .

Let $T_{n}$ denote the $n$ th term of the arithmetic series and $G_{n}$ the $n$ th term of the geometric series. The given information is

T_{4} = G_{5},\ \ T_{9} = G_{8},\ \ T_{12} = G_{15}

i.e.

a + 3d = br^{4}\quad\quad(1)

a + 8d = br^{7}\quad\quad(2)

a + 11d = br^{14}\quad\quad(3)

Subtracting $(1)$ from $(2)$ , and $(2)$ from $(3)$ ,

5d = br^{4}\left( r^{3} - 1 \right)\quad\quad(4)

3d = br^{7}\left( r^{7} - 1 \right)\quad\quad(5)

Since $d \neq 0$ , dividing $(4)$ by $(5)$ eliminates $d$ :

\frac{5}{3} = \frac{br^{4}\left( r^{3} - 1 \right)}{br^{7}\left( r^{7} - 1 \right)} = \frac{r^{3} - 1}{r^{3}\left( r^{7} - 1 \right)}

Cross-multiplying,

5r^{3}\left( r^{7} - 1 \right) = 3\left( r^{3} - 1 \right)

5r^{10} - 5r^{3} = 3r^{3} - 3

5r^{10} - 8r^{3} + 3 = 0\ \ \text{(shown)}

Note: $r = 1$ satisfies this equation exactly, but it is excluded by the given condition $|r| < 1$ .

By GC, the only root with $|r| < 1$ is

r \approx 0.74

To check your answer:

5(0.7404)^{10} - 8(0.7404)^{3} + 3 \approx 0

, using the unrounded GC value. ✓

$\therefore$ $r \approx 0.74$ (2 d.p.).

Part (ii)

3 marks

Approach

“The terms after, but not including, the $n$ th term” is the sum of the geometric series from the $(n + 1)$ th term onward. Since $|r| < 1$ , this tail sum converges, and it equals the $(n + 1)$ th term divided by $1 - r$ .

The $(n + 1)$ th term of the geometric series is $br^{n}$ , so the sum of all terms after the $n$ th term is the sum to infinity of a geometric series with first term $br^{n}$ and common ratio $r$ :

\sum_{k = n + 1}^{\infty}G_{k} = \frac{br^{n}}{1 - r}

Using $r = 0.74$ ,

\frac{br^{n}}{1 - r} = \frac{b(0.74)^{n}}{1 - 0.74} = \frac{b(0.74)^{n}}{0.26}

Simplifying the constant,

\frac{1}{0.26} = \frac{50}{13}

\therefore\ \sum_{k = n + 1}^{\infty}G_{k} = \frac{50b(0.74)^{n}}{13}

2016 H2 Maths Paper 1, Question 4

Part (i)

Part (ii)

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