C2 June 2008 Q6
6. A geometric series has first term 5 and common ratio \(\dfrac{4}{5}\).
Calculate
Given that the sum to \(k\) terms of the series is greater than 24.95,
| Scheme | Marks |
|---|---|
| \(T_{20} = 5\times\left(\dfrac{4}{5}\right)^{19} = 0.072\) (Accept awrt) Allow \(5\times\dfrac{4}{5}^{19}\) for M1 | M1 A1 |
| (2) |
Notes
(a) and (b): Correct answer without working scores both marks.
M: Requires use of the correct formula \(ar^{n-1}\).
| Scheme | Marks |
|---|---|
| \(S_\infty = \dfrac{5}{1 - 0.8} = 25\) | M1 A1 |
| (2) |
Notes
(a) and (b): Correct answer without working scores both marks.
M: Requires use of the correct formula \(\dfrac{a}{1 - r}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{5(1 - 0.8^k)}{1 - 0.8} \gt 24.95\) (Allow with = or <) | M1 |
| \(1 - 0.8^k \gt 0.998\) (or equiv., see below) (Allow with = or <) | A1 |
| \(k\log 0.8 \lt \log 0.002\) or \(k \gt \log_{0.8} 0.002\) (Allow with = or <) | M1 |
| \(k \gt \dfrac{\log 0.002}{\log 0.8}\) (*) | A1cso |
| (4) |
Notes
1st M: The sum may have already been ‘manipulated’ (perhaps wrongly), but this mark can still be allowed.
1st A: A ‘numerically correct’ version that has dealt with \((1 - 0.8)\) denominator,
e.g. \(1 - \left(\dfrac{4}{5}\right)^k \gt 0.998,\quad 5(1 - 0.8^k) \gt 4.99,\quad 25(1 - 0.8^k) \gt 24.95,\quad 5 - 5(0.8^k) \gt 4.99\). In any of these, \(\dfrac{4}{5}\) instead of 0.8 is fine, and condone \(\dfrac{4}{5}^k\) if correctly treated later.
2nd M: Introducing logs and using laws of logs correctly (this must include dealing with the power \(k\) so that \(\log p^k = k\log p\)). (corrected from the printed mark scheme, which shows \(p^k = k\log p\))
2nd A: An incorrect statement (including equalities) at any stage in the working loses this mark (this is often identifiable at the stage \(k\log 0.8 \gt \log 0.002\)).
(So a fully correct method with inequalities is required.)
| Scheme | Marks |
|---|---|
| \(k = 28\) (Must be this integer value) Not \(k \gt 27\), or \(k \lt 28\), or \(k \gt 28\) | B1 |
| (1) | |
| 9 |