OCR H240/02 2021 November — Question 3 6 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind term or common difference
DifficultyModerate -0.5 This is a straightforward two-equation system using standard arithmetic sequence formulas (nth term and sum). Students need to recall a + 14d = 88 and S₁₀ = 5(2a + 9d) = 310, then solve simultaneously. It's slightly easier than average because it's purely procedural with no conceptual challenges or problem-solving required.
Spec1.04h Arithmetic sequences: nth term and sum formulae

3 The 15th term of an arithmetic sequence is 88. The sum of the first 10 terms is 310 .
Determine the first term and the common difference.

Question 3:
AnswerMarks Guidance
AnswerMark Guidance
\(a + 14d = 88\) (i)M1, A1 M1 for one error, eg \(a + 9d = 88\)
\(\frac{10}{2}(2a + 9d) = 310\) (ii)M1, A1 M1 for \(\frac{15}{2}(2a+14d)=310\) or one error, eg \(5(2a+18d)=310\)
Substitute from (i) into (ii): \(\frac{10}{2}(2(88-14d)+9d)=310\), \(176-19d=62\)M1 Attempt substitution, elimination or verify solutions found BC. Condone one arithmetical error. SC for last 2 marks: correct answers, substitution not seen: B1B0, dep M1M1
\(d=6,\ a=4\)A1 cao
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| $a + 14d = 88$ (i) | **M1, A1** | M1 for one error, eg $a + 9d = 88$ |
| $\frac{10}{2}(2a + 9d) = 310$ (ii) | **M1, A1** | M1 for $\frac{15}{2}(2a+14d)=310$ or one error, eg $5(2a+18d)=310$ |
| Substitute from (i) into (ii): $\frac{10}{2}(2(88-14d)+9d)=310$, $176-19d=62$ | **M1** | Attempt substitution, elimination or verify solutions found BC. Condone one arithmetical error. SC for last 2 marks: correct answers, substitution not seen: B1B0, dep M1M1 |
| $d=6,\ a=4$ | **A1** | cao |

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3 The 15th term of an arithmetic sequence is 88. The sum of the first 10 terms is 310 .\\
Determine the first term and the common difference.

\hfill \mbox{\textit{OCR H240/02 2021 Q3 [6]}}