C2 June 2013 Q1
1. The first three terms of a geometric series are\[18,\ 12 \text{ and } p\]respectively, where \(p\) is a constant.
Find
| Scheme | Marks |
|---|---|
| \(\{r =\}\ \dfrac{2}{3}\) | B1 |
| (1) |
Notes
B1: Accept \(\dfrac{12}{18}\), \(0.\dot{6}\) or 0.6 recurring, or even 0.667 (3sf) but not 0.6 or 0.67
| Scheme | Marks |
|---|---|
| \(\{p =\}\ 8\) | B1 cao |
| (1) |
Notes
B1: accept 8 only
| Scheme | Marks |
|---|---|
| \(\{S_{15} =\}\ \dfrac{18\left(1 - \left(\frac{2}{3}\right)^{15}\right)}{1 - \frac{2}{3}}\) | M1 |
| \(\{S_{15} = 53.87668\ldots\} \Rightarrow S_{15} =\) awrt 53.877 | A1 |
| (2) | |
| [4] |
Notes
M1: Applies this formula \(S_{15} = \dfrac{18\left(1 - (\text{their } r)^{15}\right)}{1 - (\text{their } r)}\), can be implied by their answer. For this mark they may use any value for \(r\) except \(r = 1\) or \(r = 0\) (even 3/2 or -6 may be used)
A1: Answers which round to 53.877
Alternative method for (c)
M1: (Adding terms is an unlikely method for this question) Need to see 15 terms listed as 18+12+……0.06165877 or can be implied by correct answer
A1: awrt 53.877
Answer only : 53.9 is M0A0 with no working, but 53.877 with no working is M1A1