| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find common ratio from terms |
| Difficulty | Standard +0.3 This is a standard geometric series question requiring systematic application of formulas (ar^n for terms, sum to infinity). Part (i) involves solving simultaneous equations with ar and ar^4, part (ii) is direct formula application, and part (iii) requires algebraic manipulation to show a given result. Slightly above average difficulty due to the three-part structure and the 'show that' proof in part (iii), but all techniques are routine for C2 level. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(ar = -48,\ ar^4 = 6\) | B1 | |
| \(r^3 = \frac{6}{-48} = -\frac{1}{8}\) | M1 | |
| \(r = \sqrt[3]{-\frac{1}{8}} = -\frac{1}{2}\) | A1 | |
| \(a = \frac{-48}{-\frac{1}{2}} = 96\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(= \frac{96}{1-(-\frac{1}{2})} = 64\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(S_n = \frac{96[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})} = 64\left[1 - \left(-\frac{1}{2}\right)^n\right]\) | M1 A1 | |
| \(S_\infty - S_n = 64 - 64\left[1 - \left(-\frac{1}{2}\right)^n\right]\) | M1 | |
| \(= 64\left(-\frac{1}{2}\right)^n = 2^6 \times (-1)^n \times 2^{-n} = (-1)^n \times 2^{6-n}\) | M1 | |
| difference is magnitude, \(\therefore = 2^{6-n}\) | A1 | (11) |
| Total | (72) |
# Question 9:
## Part (i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $ar = -48,\ ar^4 = 6$ | B1 | |
| $r^3 = \frac{6}{-48} = -\frac{1}{8}$ | M1 | |
| $r = \sqrt[3]{-\frac{1}{8}} = -\frac{1}{2}$ | A1 | |
| $a = \frac{-48}{-\frac{1}{2}} = 96$ | A1 | |
## Part (ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $= \frac{96}{1-(-\frac{1}{2})} = 64$ | M1 A1 | |
## Part (iii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $S_n = \frac{96[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})} = 64\left[1 - \left(-\frac{1}{2}\right)^n\right]$ | M1 A1 | |
| $S_\infty - S_n = 64 - 64\left[1 - \left(-\frac{1}{2}\right)^n\right]$ | M1 | |
| $= 64\left(-\frac{1}{2}\right)^n = 2^6 \times (-1)^n \times 2^{-n} = (-1)^n \times 2^{6-n}$ | M1 | |
| difference is magnitude, $\therefore = 2^{6-n}$ | A1 | **(11)** |
| | Total | **(72)** |9. The second and fifth terms of a geometric series are - 48 and 6 respectively.\\
(i) Find the first term and the common ratio of the series.\\
(ii) Find the sum to infinity of the series.\\
(iii) Show that the difference between the sum of the first $n$ terms of the series and its sum to infinity is given by $2 ^ { 6 - n }$.
\hfill \mbox{\textit{OCR C2 Q9 [11]}}