| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.8 This is a straightforward geometric progression question requiring only direct application of standard formulas. Part (i) is immediate division, part (ii) uses the nth term formula with given values, and part (iii) applies the sum formula. All steps are routine recall with no problem-solving or conceptual challenges, making it easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = -2\) | B1 | State \(-2\); not \(^{-6}/_{3}\) as final answer; no need to see \(r = ...\), condone other variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt \(u_{11}\) | M1 | Must use correct formula with \(a=3\) and \(r=-2\), or their \(r\) from (i); allow M1 for \(3 \times -2^{10}\); using \(r=2\) is M0 unless that was their value in (i); allow M1 for listing terms as far as \(u_{11}\) |
| \(3 \times (-2)^{10} = 3072\) | A1 | CWO; allow A1 BOD for \(3 \times -2^{10} = 3072\); if listing terms, need to indicate 3072 is the required value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt \(S_{20}\) | M1 | Must use correct formula with \(a=3\) and \(r=-2\), or their \(r\) from (i); allow M1 for correct formula but with no brackets around the \(-2\); allow M1 for attempting to sum first 20 terms; allow M1 for \(\frac{3(1+2^{20})}{1+2}\) as long as correct general formula also seen |
| \(\frac{3(1-(-2)^{20})}{1-(-2)} = -1048575\) | A1 | Obtain \(-1048575\); could also come from manually summing terms; NB \(\frac{3(1--2^{20})}{1--2}\) gives \(1048577\) |
## Question 1:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = -2$ | B1 | State $-2$; not $^{-6}/_{3}$ as final answer; no need to see $r = ...$, condone other variables |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt $u_{11}$ | M1 | Must use correct formula with $a=3$ and $r=-2$, or their $r$ from (i); allow M1 for $3 \times -2^{10}$; using $r=2$ is M0 unless that was their value in (i); allow M1 for listing terms as far as $u_{11}$ |
| $3 \times (-2)^{10} = 3072$ | A1 | CWO; allow A1 BOD for $3 \times -2^{10} = 3072$; if listing terms, need to indicate 3072 is the required value |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt $S_{20}$ | M1 | Must use correct formula with $a=3$ and $r=-2$, or their $r$ from (i); allow M1 for correct formula but with no brackets around the $-2$; allow M1 for attempting to sum first 20 terms; allow M1 for $\frac{3(1+2^{20})}{1+2}$ as long as correct general formula also seen |
| $\frac{3(1-(-2)^{20})}{1-(-2)} = -1048575$ | A1 | Obtain $-1048575$; could also come from manually summing terms; **NB** $\frac{3(1--2^{20})}{1--2}$ gives $1048577$ |
---1 A geometric progression has first term 3 and second term - 6 .\\
(i) State the value of the common ratio.\\
(ii) Find the value of the eleventh term.\\
(iii) Find the sum of the first twenty terms.
\hfill \mbox{\textit{OCR C2 2015 Q1 [5]}}