June 2025 Paper 1 Q9
9
(a) A geometric series, \(S\), has second term 60
The common ratio of \(S\) is 0.2
Find the exact value of the sum of the first five terms of \(S\)
[4 marks](b) A different geometric series, \(T\), has second term 60 and positive common ratio \(r\)
The sum to infinity of \(T\) is \(T_{\infty}\)
(i) Show that\[T_{\infty} = \frac{60}{r - r^2}\] [2 marks]
(ii) Find the maximum value of \(r - r^2\) [2 marks]
(iii) Hence find the range of possible values of \(T_{\infty}\)
Fully justify your answer.
[2 marks]| Scheme | Marks | AO |
|---|---|---|
| Uses \(0.2a = 60\) PI by \(a = 300\) | M1 | 1.1a |
| Deduces \(a = 300\) PI by the use of \(\dfrac{60}{0.2}\) | R1 | 2.2a |
| Uses a complete method to find the sum of the five terms. For example Uses \(S_n = \dfrac{a\left(1 - r^n\right)}{1 - r}\) with their \(a \neq 60\), \(r = 0.2\) and \(n = 5\) Or Uses \(S_n = \dfrac{a\left(1 - r^n\right)}{1 - r}\) with \(a = 60\), \(r = 0.2\) and \(n = 4\) and adds their first term Or Uses their \(a = 300\) and \(r = 0.2\) to calculate all five terms and finds their sum 300 + 60 + 12 + 2.4 + 0.48 | M1 | 3.1a |
| Obtains \(\dfrac{9372}{25}\) OE | A1 | 1.1b |
| (4) |
Typical solution
\[0.2a = 60\]\[a = 300\]\[S_5 = \frac{300\left(1 - 0.2^5\right)}{1 - 0.2}\]\[= 374.88\]| Scheme | Marks | AO |
|---|---|---|
| (i) States \(\left(T_{\infty} =\right)\dfrac{a}{1 - r}\) and \(ar = 60\) or \(a = \dfrac{60}{r}\) Or States \(\left(T_{\infty} =\right)\dfrac{60/r}{1 - r}\) | M1 | 3.1a |
| Completes a reasoned argument to show \(T_{\infty} = \dfrac{60}{r - r^2}\) Must start from \(\left(T_{\infty} =\right)\dfrac{a}{1 - r}\) and \(ar = 60\) or \(a = \dfrac{60}{r}\) AG | R1 | 2.1 |
| (2) | ||
| (ii) Deduces \((r =)\,\dfrac{1}{2}\) | M1 | 2.2a |
| Deduces maximum value of \(r - r^2 = \dfrac{1}{4}\) | R1 | 1.1b |
| (2) | ||
| (iii) Obtains \(\dfrac{60}{\text{their } r - r^2}\) provided their \(r - r^2 \neq 0\) PI by 240 | M1 | 2.2a |
| Deduces \(T_{\infty} \geqslant 240\) | R1 | 2.2a |
| (2) | ||
| (10 marks) |
Typical solution
(b)(i)
\[ar = 60 \Rightarrow a = \frac{60}{r}\]\[T_{\infty} = \frac{a}{1 - r} = \frac{60}{r(1 - r)} = \frac{60}{r - r^2}\](b)(ii)
\(r - r^2\) is at its maximum when \(r = \dfrac{1}{2}\)
Maximum value of \(r - r^2 = \dfrac{1}{4}\)
(b)(iii)
\[\text{Min } T_{\infty} = \frac{60}{\frac{1}{4}} = 240\]Hence \(T_{\infty} \geqslant 240\)