OCR MEI C2 — Question 3 6 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeReal-world AP: find term or total
DifficultyEasy -1.2 This is a straightforward application of standard arithmetic sequence formulas (nth term and sum) with clear real-world context. Part (i) requires direct substitution into the nth term formula, while part (ii) involves solving a quadratic inequality from the sum formula—both are routine textbook exercises requiring only recall and basic algebraic manipulation, making this easier than average.
Spec1.04h Arithmetic sequences: nth term and sum formulae

3 On his \(1 ^ { \text {st } }\) birthday, John was given \(\pounds 5\) by his Uncle Fred. On each succeeding birthday, Uncle Fred gave a sum of money that was \(\pounds 3\) more than the amount he gave on the last birthday.
  1. How much did Uncle Fred give John on his \(8 { } ^ { \text {th } }\) birthday?
  2. On what birthday did the gift from Uncle Fred result in the total sum given on all birthdays exceeding £200?

Question 3:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
A.P.: \(5 + 8 + 11\ldots\)M1
\(\Rightarrow a_8 = 5 + 7\times 3 = £26\)A1 [2]
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
We require \(S_n > 200\) so solve \(S_n = 200\)M1
\(S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(10 + (n-1)3) = 200\)
\(\Rightarrow 10n + 3n(n-1) = 400 \Rightarrow 3n^2 + 7n - 400 = 0\)A1
\(\Rightarrow n = \frac{-7 \pm \sqrt{49 + 4800}}{6} = \frac{-7 \pm 69.6}{6} = 10.4\)M1
So minimum time is 11 years.A1 [4]
## Question 3:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| A.P.: $5 + 8 + 11\ldots$ | M1 | |
| $\Rightarrow a_8 = 5 + 7\times 3 = £26$ | A1 | **[2]** |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| We require $S_n > 200$ so solve $S_n = 200$ | M1 | |
| $S_n = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(10 + (n-1)3) = 200$ | | |
| $\Rightarrow 10n + 3n(n-1) = 400 \Rightarrow 3n^2 + 7n - 400 = 0$ | A1 | |
| $\Rightarrow n = \frac{-7 \pm \sqrt{49 + 4800}}{6} = \frac{-7 \pm 69.6}{6} = 10.4$ | M1 | |
| So minimum time is 11 years. | A1 | **[4]** |

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3 On his $1 ^ { \text {st } }$ birthday, John was given $\pounds 5$ by his Uncle Fred. On each succeeding birthday, Uncle Fred gave a sum of money that was $\pounds 3$ more than the amount he gave on the last birthday.\\
(i) How much did Uncle Fred give John on his $8 { } ^ { \text {th } }$ birthday?\\
(ii) On what birthday did the gift from Uncle Fred result in the total sum given on all birthdays exceeding £200?

\hfill \mbox{\textit{OCR MEI C2  Q3 [6]}}