C2 June 2012 Q9
9.A geometric series is \(a + ar + ar^2 + \ldots\)
The third and fifth terms of a geometric series are 5.4 and 1.944 respectively and all the terms in the series are positive.
For this series find,
| Scheme | Marks |
|---|---|
| \((S_n =)\ a + ar + (ar^2) + \ldots + ar^{n-1}\) and \(rS_n = ar + ar^2 + (ar^3) \ldots + ar^n\) | M1 |
| \(S_n - rS_n = a - ar^n\) | M1 |
| \(S_n(1 - r) = a(1 - r^n)\) | dM1 |
| And so result \(S_n = \dfrac{a(1 - r^n)}{(1 - r)}\) * | A1 |
| (4) |
Notes
M1: Lists both of these sums (\(S_n =\)) may be omitted, \(rS_n\) (or \(rS\)) must be stated
1st two terms must be correct in each series. Last term must be \(ar^{n-1}\) or \(ar^n\) in first series and the corresponding \(ar^n\) or \(ar^{n+1}\) in second series. Must be \(n\) and not a number. Reference made to other terms e.g. space or dots to indicate missing terms
M1: Subtracts series for \(rS\) from series for \(S\) (or other way round) to give RHS \(= \pm(a - ar^n)\). This may have been obtained by following a pattern. If wrong power stated on line 1 M0 here. (Ignore LHS)M0M0M0A0
dM1: Factorises both sides correctly– must follow from a previous M1 (It is possible to obtain M0M1M1A0 or M1M0M1A0) A1: completes the proof with no errors seen
Special Case
No errors seen: First line absolutely correct, omission of second line, third and fourth lines correct: M1M0M1A1
See next sheet of common errors.
Refer any attempts involving sigma notation, or any proofs by induction to team leader.
Also attempts which begin with the answer and work backwards.
| Scheme | Marks | |
|---|---|---|
| Method 1 | Method 2 | |
| Divides one term by other (either way) to give \(r^2 = \ldots\) then square roots to give \(r =\) | Or: (Method 2) Finds geometric mean i.e 3.24 and divides one term by 3.24 or 3.24 by one term | M1 |
| \(r^2 = \dfrac{1.944}{5.4},\quad r = 0.6\) (ignore \(-0.6\)) | \(r = 0.6\) (ignore \(-0.6\)) | A1 |
| (2) | ||
Notes
M1: Deduces \(r^2\) by dividing either term by other and attempts square root
A1: any correct equivalent for \(r\) e.g. 3/5 Answer only is 2/2
(Method 2) Those who find fourth term must use \(\sqrt{ab}\) and not \(\tfrac{1}{2}(a + b)\) then must use it in a division with given term to obtain \(r =\)
| Scheme | Marks |
|---|---|
| Uses \(5.4 \div r^2\) or \(1.944 \div r^4\), to give \(a =\) \(a = 15\) | M1, A1ft |
| (2) |
Notes
M1: May be done in two steps or more e.g. \(5.4 \div r\) then divided by \(r\) again
A1ft: follow through their value of \(r\). Just \(a = 15\) with no wrong working implies M1A1
| Scheme | Marks |
|---|---|
| Uses \(S = \dfrac{15}{1 - 0.6}\), to obtain 37.5 | M1A1, A1 |
| (3) | |
| 11 marks |
Notes
M1: States sum to infinity formula with values of \(a\) and \(r\) found earlier, provided \(|r| \lt 1\)
A1: uses 15 and 0.6 (or 3/5) (This is not a ft mark) A1: 37.5 or exact equivalent
Common errors
(i) Fraction inverted in (b) \(r^2 = \dfrac{5.4}{1.944}\) and \(r = 1\tfrac{2}{3}\), then correct ft gives M1A0 M1 A1ft M0A0A0 i.e. 3/7
(ii) Uses \(r = 0.36\): (b)M0A0 (c)M1A1ft (d) M1A0A0 i.e. 3/7
(iii) Uses \(ar^3 = 5.4,\ ar^5 = 1.944\) Likely to have (b)M1A1 (c)M0A0 (d) M1A0A0 i.e.3/7