CAIE P1 2018 November — Question 5 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionNovember
Marks6
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Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind term or common difference
DifficultyModerate -0.5 This is a straightforward two-equation system using standard arithmetic progression formulas (nth term and sum). It requires only direct substitution and basic algebra, making it slightly easier than average but not trivial due to the simultaneous equation solving required.
Spec1.04h Arithmetic sequences: nth term and sum formulae
5 In an arithmetic progression the first term is \(a\) and the common difference is 3 . The \(n\)th term is 94 and the sum of the first \(n\) terms is 1420 . Find \(n\) and \(a\).
Question 5:
AnswerMarks Guidance
\(a + (n-1)3 = 94\)B1
\(\frac{n}{2}[2a+(n-1)3] = 1420\) OR \(\frac{n}{2}[a+94] = 1420\)B1
Attempt elimination of \(a\) or \(n\)M1
\(3n^2 - 191n + 2840 (=0)\) OR \(a^2 - 3a - 598 (=0)\)A1 3-term quadratic (not necessarily all on same side)
\(n = 40\) (only)A1
\(a = -23\) (only)A1 Award 5/6 if 2nd pair of solutions \((71/3, 26)\) given in addition or as only answer 
## Question 5:

| $a + (n-1)3 = 94$ | B1 | |
|---|---|---|
| $\frac{n}{2}[2a+(n-1)3] = 1420$ OR $\frac{n}{2}[a+94] = 1420$ | B1 | |
| Attempt elimination of $a$ or $n$ | M1 | |
| $3n^2 - 191n + 2840 (=0)$ OR $a^2 - 3a - 598 (=0)$ | A1 | 3-term quadratic (not necessarily all on same side) |
| $n = 40$ (only) | A1 | |
| $a = -23$ (only) | A1 | Award 5/6 if 2nd pair of solutions $(71/3, 26)$ given in addition or as only answer |

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5 In an arithmetic progression the first term is $a$ and the common difference is 3 . The $n$th term is 94 and the sum of the first $n$ terms is 1420 . Find $n$ and $a$.\\

\hfill \mbox{\textit{CAIE P1 2018 Q5 [6]}}
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