2020 H2 Maths Paper 1 Q8 — Worked Solution

Part (a)(i)

2 marks

Let the first term be $a = 4$ and the common difference $d$ . Since the 5th term is $10$ ,

a + 4d = 10

\Rightarrow 4 + 4d = 10

\Rightarrow 4d = 6

\Rightarrow d = 1.5

The 30th term is

a + 29d = 4 + 29(1.5) = 4 + 43.5 = 47.5

To check your answer: the 5th term is

4 + 4(1.5) = 4 + 6 = 10

, matching the given value. ✓

$\therefore$ the 30th term is $47.5$ .

Part (a)(ii)

3 marks

The sum of the first $n$ terms of an AP is $S_{n} = \frac{n}{2}\left( 2a + (n - 1)d \right)$ . The sum from the 21st to the 50th term inclusive is $S_{50} - S_{20}$ :

S_{50} = \frac{50}{2}\left( 2(4) + 49(1.5) \right) = 25(8 + 73.5) = 25(81.5) = 2037.5

S_{20} = \frac{20}{2}\left( 2(4) + 19(1.5) \right) = 10(8 + 28.5) = 10(36.5) = 365

S_{50} - S_{20} = 2037.5 - 365 = 1672.5

To check your answer: the 21st and 50th terms are

4 + 20(1.5) = 34

and

4 + 49(1.5) = 77.5

; summing these

30

terms as

\frac{30}{2}(34 + 77.5) = 15(111.5) = 1672.5

matches the difference of sums above. ✓

$\therefore$ the sum from the 21st to the 50th term inclusive is $1672.5$ .

Part (b)(i)

2 marks

Let the first term be $a = 4$ and common ratio $r > 0$ . Since the 5th term is $1.6384$ ,

ar^{4} = 1.6384

\Rightarrow 4r^{4} = 1.6384

\Rightarrow r^{4} = 0.4096

Recognising $0.4096 = {0.8}^{4}$ (since ${0.8}^{2} = 0.64$ and ${0.64}^{2} = 0.4096$ ), and taking the positive root,

r = 0.8

Since $|r| = 0.8 < 1$ , the sum to infinity exists:

S_{\infty} = \frac{a}{1 - r} = \frac{4}{1 - 0.8} = \frac{4}{0.2} = 20

To check your answer: the 5th term is

4(0.8)^{4} = 4(0.4096) = 1.6384

, matching the given value. ✓

$\therefore$ $S_{\infty} = 20$ .

Part (b)(ii)

5 marks

Approach

We first write $S_{n}$ in terms of $r = 0.8$ and manipulate the given inequality $S_{n} > 19.6$ algebraically to reach ${0.8}^{n} < 0.02$ , tracking the direction of each inequality carefully — multiplying by $- 1$ reverses it. Taking logarithms then isolates $n$ ; since $\ln 0.8 < 0$ , dividing by it reverses the inequality again.

The sum of the first $n$ terms is

S_{n} = \frac{a\left( 1 - r^{n} \right)}{1 - r} = \frac{4\left( 1 - {0.8}^{n} \right)}{0.2} = 20\left( 1 - {0.8}^{n} \right)

Given $S_{n} > 19.6$ ,

20\left( 1 - {0.8}^{n} \right) > 19.6

\Rightarrow 1 - {0.8}^{n} > 0.98

\Rightarrow - {0.8}^{n} > - 0.02

\Rightarrow {0.8}^{n} < 0.02\ \ \text{(multiplying by } - 1\text{, reversing the inequality)}

$\ \$ (shown)

Taking logarithms of both sides (noting $\ln 0.8 < 0$ , so the inequality reverses),

{0.8}^{n} < 0.02

\Rightarrow n\ln 0.8 < \ln 0.02

\Rightarrow n > \frac{\ln 0.02}{\ln 0.8}

To check your answer:

\frac{\ln 0.02}{\ln 0.8} \approx \frac{- 3.912}{- 0.2231} \approx 17.53

; testing

n = 17

:

S_{17} = 20\left( 1 - {0.8}^{17} \right) \approx 19.55 < 19.6

(fails), and

n = 18

:

S_{18} = 20\left( 1 - {0.8}^{18} \right) \approx 19.64 > 19.6

(succeeds). ✓

Since $n$ must be an integer and $n > 17.53$ , the smallest possible value is $n = 18$ .

$\therefore$ the smallest possible value of $n$ is $18$ .

2020 H2 Maths Paper 1, Question 8

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

This is how we mark, and how we teach.