C2 January 2008 Q2
2. The fourth term of a geometric series is 10 and the seventh term of the series is 80.
For this series, find
| Scheme | Marks |
|---|---|
| Complete method, using terms of form \(ar^k\), to find \(r\) [e.g. Dividing \(ar^6 = 80\) by \(ar^3 = 10\) to find \(r\); \(\;r^6 - r^3 = 8\) is M0] | M1 |
| \(r = 2\) | A1 |
| (2) |
Notes
M1: Condone errors in powers, e.g. \(ar^4 = 10\) and/or \(ar^7 = 80\),
A1: For \(r = 2\), allow even if \(ar^4 = 10\) and \(ar^7 = 80\) used (just these)
(M mark can be implied from numerical work, if used correctly)
In (a) and (b) correct answer, with no working, allow both marks.
| Scheme | Marks |
|---|---|
| Complete method for finding \(a\) [e.g. Substituting value for \(r\) into equation of form \(ar^k = 10\) or 80 and finding a value for \(a\).] | M1 |
| \((8a = 10)\qquad a = \dfrac{5}{4} = 1\dfrac{1}{4}\) (equivalent single fraction or 1.25) | A1 |
| (2) |
Notes
M1: Allow for numerical approach: e.g. \(\dfrac{10}{r_c^{\,3}} \ \leftarrow\ \dfrac{10}{r_c^{\,2}} \ \leftarrow\ \dfrac{10}{r_c} \leftarrow 10\)
In (a) and (b) correct answer, with no working, allow both marks.
| Scheme | Marks |
|---|---|
| Substituting their values of \(a\) and \(r\) into correct formula for sum. | M1 |
| \(S = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{5}{4}(2^{20} - 1)\) (= 1310718.75) 1 310 719 (only this) | A1 |
| (2) | |
| [6] |
Notes
Attempt 20 terms of series and add is M1 (correct last term 655360)
If formula not quoted, errors in applying their \(a\) and/or \(r\) is M0
Allow full marks for correct answer with no working seen.