C2 June 2015 Q5
5.
Find
Find the smallest value of \(n\) for which the sum of the first \(n\) terms of the series exceeds 290
(4)| Scheme | Marks |
|---|---|
| Mark (a) and (b) together | |
| (a) \(a + ar = 34\) or \(\dfrac{a(1 - r^2)}{(1 - r)} = 34\) or \(\dfrac{a(r^2 - 1)}{(r - 1)} = 34\); \(\dfrac{a}{1 - r} = 162\) | B1; B1 |
| (Way 1) Eliminate \(a\) to give \((1 + r)(1 - r) = \dfrac{17}{81}\) or \(1 - r^2 = \dfrac{34}{162}\).. (not a cubic) | aM1 |
| (and so \(r^2 = \tfrac{64}{81}\) and) \(r = \tfrac{8}{9}\) only | aA1 (4) |
| (b) Substitute their \(r = \tfrac{8}{9}\) \((0 \lt r \lt 1)\) to give \(a =\) | bM1 |
| \(a = 18\) | bA1 (2) |
Notes
(Way 2) Part (b) first
| Scheme | Marks |
|---|---|
| Eliminate \(r\) to give \(\dfrac{34 - a}{a} = 1 - \dfrac{a}{162}\) | bM1 |
| gives \(a = 18\) or 306 and rejects 306 to give \(a = 18\) | bA1 |
| Then part (a) again: Substitute a = 18 to give \(r =\) | aM1 |
| \(r = \tfrac{8}{9}\) | aA1 |
(i) (a) B1: Writes a correct equation connecting \(a\) and \(r\) and 34 (allow equivalent equations – may be implied)
B1: Writes a correct equation connecting \(a\) and \(r\) and 162 (allow equivalent equation – may be implied)
Way 1: aM1: Eliminates \(a\) correctly for these two equations to give \((1 + r)(1 - r) = \dfrac{17}{81}\) or \((1 + r)(1 - r) = \dfrac{34}{162}\) or equivalent – not a cubic – should have factorized \((1 - r)\) to give a correct quadratic
aA1: Correct value for \(r\). Accept 0.8 recurring or 8/9 (not 0.889) Must only have positive value.
bM1: Substitutes their \(r\) \((0 \lt r \lt 1)\) into a correct formula to give value for \(a\). Can be implied by \(a = 18\)
bA1: must be 18 (not answers which round to 18)
Way 2: Finds \(a\) first - B1, B1: As before then award the (b) M and A marks before the (a) M and A marks
bM1: Eliminates \(r\) correctly to give \(\dfrac{34 - a}{a} = 1 - \dfrac{a}{162}\) or \(a^2 - 324a + 5508 = 0\) or equivalent
bA1: Correct value for \(a\) so \(a = 18\) only. (Only award after 306 has been rejected)
aM1: Substitutes their 18 to give \(r =\)
aA1: \(r = \tfrac{8}{9}\) only
| Scheme | Marks |
|---|---|
| \(\dfrac{42\left(1 - \left(\frac{6}{7}\right)^n\right)}{1 - \frac{6}{7}} \gt 290\) (For trial and improvement approach see notes below) | M1 |
| to obtain So \(\left(\tfrac{6}{7}\right)^n \lt \left(\tfrac{4}{294}\right)\) or equivalent e.g. \(\left(\tfrac{7}{6}\right)^n \gt \left(\tfrac{294}{4}\right)\) or \(\left(\tfrac{6}{7}\right)^n \lt \left(\tfrac{2}{147}\right)\) | A1 |
| So \(n \gt \dfrac{\log\text{"}\left(\frac{4}{294}\right)\text{"}}{\log\left(\frac{6}{7}\right)}\) or \(\log_{\frac{6}{7}}\text{"}\left(\tfrac{4}{294}\right)\text{"}\) or equivalent but must be log of positive quantity | M1 |
| (i.e. \(n \gt 27.9\) ) so \(n = 28\) | A1 |
| (4) | |
| [10] |
Notes
(ii) M1: Allow \(n\) or \(n - 1\) and any symbols from “>”, “<”, or “=” etc A1 : Must be power \(n\) ( not \(n - 1\)) with any symbol
M1: Uses logs correctly on \(\left(\tfrac{6}{7}\right)^n\) or \(\left(\tfrac{7}{6}\right)^n\) not on \((36)^n\) to get as far as \(n\) Allow any symbol
A1: \(n = 28\) cso (any errors with inequalities earlier e.g. failure to reverse the inequality when dividing by the negative \(\log\left(\tfrac{6}{7}\right)\) or any contradictory statements must be penalised here) Those with equals throughout may gain this mark if they follow 27.9 by \(n\)=28. Just \(n = 28\) without mention of 27.9 is only allowed following correct inequality work.
Special case: Trial and improvement: Gives \(n = 28\) as \(S\) = awrt 290.1 (M1A1)and when \(n = 27\) \(S\) = (awrt) 289 so \(n = 28\) (M1A1) – \(n = 28\) with no working is M1A0M0A0 and insufficient accuracy is M1A0M1A0
Uses nth term instead of sum of n terms – over simplified – do not treat as misread – award 0/4