C2 January 2005 Q6
6. The second and fourth terms of a geometric series are 7.2 and 5.832 respectively.
The common ratio of the series is positive.
For this series, find
(a) the common ratio, (2)
(b) the first term, (2)
(c) the sum of the first 50 terms, giving your answer to 3 decimal places, (2)
(d) the difference between the sum to infinity and the sum of the first 50 terms, giving your answer to 3 decimal places. (2)
| Scheme | Marks |
|---|---|
| \(ar = 7.2, ar^3 = 5.832 \Rightarrow r^2 = \dfrac{5.832}{7.2}\ (= 0.81)\) | M1 |
| \(r = 0.9\) | A1 |
| (2) |
Notes
M1 for full method \(\to r^2\) or \(r\)
N.B. \(ar^2 = 7.2, ar^4 = 5.832 \to r = 0.9\) scores M1A1 in part (a) but probably M0A0 in (b).
| Scheme | Marks |
|---|---|
| \(a = \dfrac{7.2}{(a)}, = \underline{8}\) | M1, A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(s_{50} = \dfrac{8\left(1 - (0.9)^{50}\right)}{1 - 0.9}\) | M1 |
| \(= \underline{79.588}\) (3 dp) | A1 c.a.o |
| (2) |
Notes
M1ft their "\(a\)", "\(r\)" in \(s_{50}\) formula
| Scheme | Marks |
|---|---|
| \(s_\infty = \dfrac{8}{1 - 0.9}\ (= 80)\) | M1 |
| \(s_\infty - s_{50} = 80 - (c) = 0.412\) (Awrt 3 dp) | A1ft |
| (2) | |
| (8 marks) |
Notes
M1ft their "\(a\)", "\(r\)" in \(s_\infty\)
A1ft for \(80 -\) their (c) i.e. ft their (c) only