C2 June 2018 Q6
6. A geometric series with common ratio \(r = -0.9\) has sum to infinity 10 000
For this series,
| Scheme | Marks |
|---|---|
| \(10\,000 = \dfrac{a}{1 - (-0.9)}\) | M1 |
| \(a = 19\,000\) | A1 |
| (2) |
Notes
M1: Correct use of formula for sum to infinity as above, or states correct formula and makes small slip such as replacing \(r\) with 0.9 instead of \(-0.9\)
A1: Correct answer
| Scheme | Marks |
|---|---|
| Use \(ar^4\) | M1 |
| \(19\,000 \times (-0.9)^4 = 12465.9\) (accept awrt 12466) | A1 |
| (2) |
Notes
M1: Correct use of formula with \(n - 1 = 4\), allow 0.9 instead of -0.9 here. Condone invisible brackets.
A1: accept awrt 12466 (even following use of 0.9) Correct answer implies M1A1 even with no method shown. Accept correct equivalents such as mixed or improper fractions
| Scheme | Marks |
|---|---|
| \(S = \dfrac{a(1 - r^{12})}{1 - r}\) or lists and adds their first twelve terms with their \(a\) | M1 |
| \(S = \dfrac{\text{"}19000\text{"}\left(1 - (-0.9)^{12}\right)}{1 - (-0.9)}\) or \(S = 10000\left(1 - (-0.9)^{12}\right)\) | A1ft |
| \(= 7176\) only | A1cso |
| (3) | |
| [7] |
Notes
M1: Correct use of formula with power 12 (or adds 12 terms) with their \(a\) (not 10000) and \(r = +0.9\) or -0.9
A1ft: Correct unsimplified with their \(a\) and with \(r = +0.9\) or -0.9 or for listing method as follows
19000 + -17100 + 15390 + -13851 + 12465.9 + -11219.31 + 10097.379 + -9087.6411 + 8178.87699 + -7360.989291 + 6624.890362 + -5962.401326 = (Do not follow through for listing method)
A1cso: 7176 only
Special case: \(S = \dfrac{a(1 - r^n)}{1 - r}\) so \(S = \dfrac{\text{"}19000\text{"}\left(1 + (0.9)^{12}\right)}{1 + (0.9)}\) is M1A0A0
Whereas \(S = \dfrac{\text{"}19000\text{"}\left(1 + (0.9)^{12}\right)}{1 + (0.9)}\) on its own with no formula quoted is M0A0A0
\(S = \dfrac{\text{"}19000\text{"}\left(1 - -0.9^{12}\right)}{1 - -0.9}\) should have M1 (bod) then final two A marks depend on whether answer is correct so if this is followed by 7176 the A1A1 should be awarded. If it is followed by 12824 then A0A0 is implied.