C2 January 2007 Q10
10. A geometric series is \(a + ar + ar^2 + \ldots\)
| Scheme | Marks |
|---|---|
| \(\{S_n =\}\ a + ar + \ldots + ar^{n-1}\) | B1 |
| \(\{rS_n =\}\ ar + ar^2 + \ldots + ar^n\) | M1 |
| \((1 - r)S_n = a(1 - r^n)\) | dM1 |
| \(S_n = \dfrac{a(1 - r^n)}{1 - r}\) (✱) | A1cso |
| (4) |
Notes
B1: \(S_n\) not required. The following must be seen: at least one + sign, \(a\), \(ar^{n-1}\) and one other intermediate term. No extra terms (usually \(ar^n\)).
M1: Multiply by \(r\); \(rS_n\) not required. At least 2 of their terms on RHS correctly multiplied by \(r\).
dM1: Subtract both sides: LHS must be \(\pm(1 - r)S_n\), RHS must be in the form \(\pm a(1 - r^{pn+q})\).
Only award this mark if the line for \(S_n = \ldots\) or the line for \(rS_n = \ldots\) contains a term of the form \(ar^{cn+d}\)
Method mark, so may contain a slip but not awarded if last term of their \(S_n\) = last term of their \(rS_n\).
A1 cso: Completion c.s.o. N.B. Answer given in question
Alternative
| \(S_n\) not required. The following must be seen: at least one + sign, \(a\), \(ar^{n-1}\) and one other intermediate term. No extra terms (usually \(ar^n\)). | B1 |
| On RHS, multiply by \(\dfrac{1 - r}{1 - r}\) Or Multiply LHS and RHS by \((1 - r)\) | M1 |
| Multiply by \((1 - r)\) convincingly (RHS) and take out factor of \(a\). Method mark, so may contain a slip. | dM1 |
| Completion c.s.o. N.B. Answer given in question | A1 cso |
| Scheme | Marks |
|---|---|
| \(a = 200,\ r = 2,\ n = 10,\quad S_{10} = \dfrac{200(1 - 2^{10})}{1 - 2}\) | M1, A1 |
| \(= 204{,}600\) | A1 |
| (3) |
Notes
M1: Substitute \(r = 2\) with \(a = 100\) or 200 and \(n = 9\) or 10 into formula for \(S_n\).
A1: \(\dfrac{200(1 - 2^{10})}{1 - 2}\) or equivalent.
A1: 204,600
10(b) Alternative method: adding 10 terms
| (i) Answer only: full marks. (M1 A1 A1) | |
| (ii) \(200 + 400 + 800 + \ldots\ \{+ 102{,}400\} = 204{,}600\) or \(100(2 + 4 + 8 + \ldots\ \{+ 1{,}024\}) = 204{,}600\) M1 for two correct terms (as above o.e.) and an indication that the sum is needed (e.g. + sign or the word sum). | M1 |
| 102,400 o.e. as final term. Can be implied by a correct final answer. | A1 |
| 204,600. | A1 |
| Scheme | Marks |
|---|---|
| \(a = \dfrac{5}{6},\ r = \dfrac{1}{3}\) | B1 |
| \(S_\infty = \dfrac{a}{1 - r}, \qquad S_\infty = \dfrac{\tfrac{5}{6}}{1 - \tfrac{1}{3}}\) | M1 |
| \(= \dfrac{5}{4}\) o.e. | A1 |
| (3) |
Notes
N.B. \(S_\infty = \tfrac{a}{1 - r}\) is in the formulae book.
B1: \(r = \dfrac{1}{3}\) seen or implied anywhere.
M1: Substitute \(a = \dfrac{5}{6}\) and their \(r\) into \(\dfrac{a}{1 - r}\). Usual rules about quoting formula.
A1: \(\dfrac{5}{4}\) o.e.
| Scheme | Marks |
|---|---|
| \(-1 \lt r \lt 1\) (or \(|r| \lt 1\)) | B1 |
| (1) | |
| (11) |
Notes
N.B. \(S_\infty = \tfrac{a}{1 - r}\) for \(|r| \lt 1\) is in the formulae book.
B1: \(-1 \lt r \lt 1\) or \(|r| \lt 1\) In words or symbols.
Take symbols if words and symbols are contradictory. Must be \(\lt\) not \(\leqslant\).