C2 January 2014 (IAL) Q4
4. The first term of a geometric series is 5 and the common ratio is 1.2
For this series find, to 1 decimal place,
(a)
(i) the 20th term,
(ii) the sum of the first 20 terms. (4)
The sum of the first \(n\) terms of the series is greater than 3000
(b) Calculate the smallest possible value of \(n\). (4)
| Scheme | Marks |
|---|---|
| (i) \(t_{20} = 5 \times 1.2^{19} = 159.7\) | M1A1 |
| (ii) \(S_{20} = \dfrac{5\left(1 - 1.2^{20}\right)}{1 - 1.2} = 933.4\) | M1A1 |
| (4) |
Notes
(i) M1: Use of \(t_n = ar^{n-1}\) A1: Cao
(ii) M1: Use of a correct sum formula with \(n = 19\) or \(n = 20\)
NB if \(n = 19\) is used and no formula is quoted, score M0
A1: Cao
| Scheme | Marks |
|---|---|
| \(\dfrac{5\left(1 - 1.2^n\right)}{1 - 1.2}\ (> \text{or} =)\ 3000\) | B1 |
| \(1.2^n > 121\) | M1 |
| \(\log 1.2^n > \log 121\) or \(n > \log_{1.2} 121\) | M1 |
| \(n > \dfrac{\log 121}{\log 1.2}\) i.e. \(n = 27\) | A1 |
| Ignore symbols e.g. ‘=’ throughout with no errors getting \(n = 27\) scores full marks | |
| In (b) Treat \(5 \times 1.2^{n-1} > 3000\) as a misread and allow the M’s if scored (gives \(n = 37\)) | |
| (4) | |
| Total 8 |
Notes
B1: Correct statement (allow ‘a’ and/or ‘r’ instead of 5 and 1.2)
M1: \(1.2^n\) (> or < or =) \(k\)
M1: Takes logs correctly
A1: cao