CAIE P1 2008 November — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2008
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind n given sum condition
DifficultyModerate -0.8 This is a straightforward application of standard arithmetic progression formulas. Given a₁=6, a₅=12, students find d=1.5 using the nth term formula, then solve S_n=90 using the sum formula to get a quadratic in n. All steps are routine with no conceptual challenges—easier than average but requires careful algebraic manipulation.
Spec1.04h Arithmetic sequences: nth term and sum formulae

3 The first term of an arithmetic progression is 6 and the fifth term is 12 . The progression has \(n\) terms and the sum of all the terms is 90 . Find the value of \(n\).

AnswerMarks Guidance
\(1^{\text{st}} \text{ term} = a = 6\)B1 Correct value of \(d\)
\(5^{\text{th}} \text{ term} = a + 4d = 12 \rightarrow d = 1.5\)M1 Use of correct formula with his \(d\)
\(S_n = \frac{n}{2}(12 + (n-1)1.5) = 90\)DM1, A1 Correct method for soln of quadratic. Co (ignore inclusion of \(n = -15\))
\(\rightarrow n^2 + 7n - 120 = 0 \rightarrow n = 8\)
$1^{\text{st}} \text{ term} = a = 6$ | B1 | Correct value of $d$

$5^{\text{th}} \text{ term} = a + 4d = 12 \rightarrow d = 1.5$ | M1 | Use of correct formula with his $d$

$S_n = \frac{n}{2}(12 + (n-1)1.5) = 90$ | DM1, A1 | Correct method for soln of quadratic. Co (ignore inclusion of $n = -15$)

$\rightarrow n^2 + 7n - 120 = 0 \rightarrow n = 8$

---
3 The first term of an arithmetic progression is 6 and the fifth term is 12 . The progression has $n$ terms and the sum of all the terms is 90 . Find the value of $n$.

\hfill \mbox{\textit{CAIE P1 2008 Q3 [4]}}