C2 January 2012 Q1
1. A geometric series has first term \(a = 360\) and common ratio \(r = \dfrac{7}{8}\)
Giving your answers to 3 significant figures where appropriate, find
(a) the 20th term of the series, (2)
(b) the sum of the first 20 terms of the series, (2)
(c) the sum to infinity of the series. (2)
| Scheme | Marks |
|---|---|
| Uses \(360 \times \left(\tfrac{7}{8}\right)^{19}\), to obtain 28.5 | M1, A1 |
| (2) |
Notes
M1: Correct use of formula with power = 19 A1: Accept 28.47, or 28.474 or indeed 28.47446075
Alternative to (a)
| Scheme | Marks |
|---|---|
| Gives all 20 terms 315, 275.6(25), 241.17(1875), … (1st 3 accurate) | M1 |
| All correct and last term as above A1: Accept 28.5, 28.47, or 28.474 or indeed 28.47446075 | A1 |
| Scheme | Marks |
|---|---|
| Uses \(S = \dfrac{360\left(1 - \left(\frac{7}{8}\right)^{20}\right)}{1 - \frac{7}{8}}\), or \(S = \dfrac{360\left(\left(\frac{7}{8}\right)^{20} - 1\right)}{\frac{7}{8} - 1}\) to obtain 2680 | M1, A1 |
| (2) |
Notes
M1: Correct use of formula with \(n = 20\) A1: Accept 2681, 2680.7, 2680.68 or 2680.679 or indeed 2680.678775 (N.B. 2680.67 or 2680.0 is A0)
Alternative to (b)
| Scheme | Marks |
|---|---|
| Gives all 20 terms 315, 275.6(25), 241.17(1875), … (1st 3 accurate) and adds | M1 |
| Sum correct A1: Accept 2680, 2681, 2680.7, 2680.68 or 2680.679 or indeed 2680.678775 | A1 |
| Scheme | Marks |
|---|---|
| Uses \(S = \dfrac{360}{1 - \frac{7}{8}}\), to obtain 2880 | M1, A1cao |
| (2) | |
| 6 |
Notes
M1: Correct use of formula A1: Accept 2880 only