| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Sum of first n terms |
| Difficulty | Moderate -0.3 This is a straightforward two-part question on arithmetic and geometric sequences requiring only standard formula application. Part (i) uses basic AP sum formula with given first term and common difference. Part (ii) involves GP sum formula, simple algebraic manipulation to reach the given inequality, and routine logarithm calculation. No problem-solving insight needed, just direct application of memorized formulas. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{30} = \frac{30}{2}(2 \times 6 + 29 \times 1.8)\) | M1 | Use \(d = 1.8\) in AP formula. Could be attempting \(S_{30}\) or \(u_{30}\). Formula must be recognisable, though not necessarily fully correct, so allow M1 for e.g. \(15(6 + 29 \times 1.8)\), \(15(12 + 14 \times 1.8)\) or even \(15(12 + 19 \times 1.8)\). Must have \(d = 1.8\) (not 1.3), \(n = 30\) and \(a = 6\) |
| Correct unsimplified \(S_{30}\) | A1 | Formula must now be fully correct. Allow for any unsimplified correct expression. If using \(\frac{1}{2}n(a+l)\) then \(l\) must be correct when substituted |
| \(= 963\) | A1 [3] | Units not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = \frac{7.8}{6} = 1.3\) | M1 | Use \(r = 1.3\) in GP formula. Could be attempting \(S_N\), \(u_N\) or even \(S_\infty\). Formula must be recognisable, though not necessarily fully correct. Must have \(r = 1.3\) (not 1.8) and \(a = 6\) |
| \(\frac{6(1-1.3^N)}{1-1.3} \leq 1800\) | A1 | Correct unsimplified \(S_N\). Formula must now be fully correct. Allow for any unsimplified correct expression |
| \(1 - 1.3^N \geq -90\) | M1 | Link sum of GP to 1800 and attempt to rearrange to \(1.3^N \leq k\). Must have used correct formula for \(S_N\) of GP. Allow \(=\), \(\geq\) or \(\leq\). Allow slips when rearranging, including with indices, so rearranging to \(7.8^N \leq k\) could get M1 |
| \(1.3^N \leq 91\) AG | A1 | Obtain given inequality. Must have correct inequality signs throughout. Correct working only, so A0 if \(6 \times 1.3^N\) becomes \(7.8^N\), even if subsequently corrected |
| \(N\log 1.3 \leq \log 91\) | M1 | Introduce logs throughout and attempt to solve equation/inequality. Must be using \(1.3^N \leq 91\), \(1.3^N = 91\) or \(1.3^N \geq 91\). M0 if no evidence of use of logarithms. M0 if invalid use of logarithms in attempt to solve |
| \(N \leq 17.19\), hence \(N = 17\) | A1 [6] | Must come from solving \(1.3^N \leq 91\) or \(1.3^N = 91\) (not incorrect inequality sign). Answer must be integer value. Answer must be equality, so A0 for \(N \leq 17\) |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{30} = \frac{30}{2}(2 \times 6 + 29 \times 1.8)$ | M1 | Use $d = 1.8$ in AP formula. Could be attempting $S_{30}$ or $u_{30}$. Formula must be recognisable, though not necessarily fully correct, so allow M1 for e.g. $15(6 + 29 \times 1.8)$, $15(12 + 14 \times 1.8)$ or even $15(12 + 19 \times 1.8)$. Must have $d = 1.8$ (not 1.3), $n = 30$ and $a = 6$ |
| Correct unsimplified $S_{30}$ | A1 | Formula must now be fully correct. Allow for any unsimplified correct expression. If using $\frac{1}{2}n(a+l)$ then $l$ must be correct when substituted |
| $= 963$ | A1 [3] | Units not required |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \frac{7.8}{6} = 1.3$ | M1 | Use $r = 1.3$ in GP formula. Could be attempting $S_N$, $u_N$ or even $S_\infty$. Formula must be recognisable, though not necessarily fully correct. Must have $r = 1.3$ (not 1.8) and $a = 6$ |
| $\frac{6(1-1.3^N)}{1-1.3} \leq 1800$ | A1 | Correct unsimplified $S_N$. Formula must now be fully correct. Allow for any unsimplified correct expression |
| $1 - 1.3^N \geq -90$ | M1 | Link sum of GP to 1800 and attempt to rearrange to $1.3^N \leq k$. Must have used correct formula for $S_N$ of GP. Allow $=$, $\geq$ or $\leq$. Allow slips when rearranging, including with indices, so rearranging to $7.8^N \leq k$ could get M1 |
| $1.3^N \leq 91$ **AG** | A1 | Obtain given inequality. Must have correct inequality signs throughout. Correct working only, so A0 if $6 \times 1.3^N$ becomes $7.8^N$, even if subsequently corrected |
| $N\log 1.3 \leq \log 91$ | M1 | Introduce logs throughout and attempt to solve equation/inequality. Must be using $1.3^N \leq 91$, $1.3^N = 91$ or $1.3^N \geq 91$. M0 if no evidence of use of logarithms. M0 if invalid use of logarithms in attempt to solve |
| $N \leq 17.19$, hence $N = 17$ | A1 [6] | Must come from solving $1.3^N \leq 91$ or $1.3^N = 91$ (not incorrect inequality sign). Answer must be integer value. Answer must be equality, so A0 for $N \leq 17$ |
---6 Sarah is carrying out a series of experiments which involve using increasing amounts of a chemical. In the first experiment she uses 6 g of the chemical and in the second experiment she uses 7.8 g of the chemical.\\
(i) Given that the amounts of the chemical used form an arithmetic progression, find the total amount of chemical used in the first 30 experiments.\\
(ii) Instead it is given that the amounts of the chemical used form a geometric progression. Sarah has a total of 1800 g of the chemical available. Show that $N$, the greatest number of experiments possible, satisfies the inequality
$$1.3 ^ { N } \leqslant 91 ,$$
and use logarithms to calculate the value of $N$.
\hfill \mbox{\textit{OCR C2 2013 Q6 [9]}}