CAIE P1 2012 November — Question 1 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2012
SessionNovember
Marks4
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TopicArithmetic Sequences and Series
TypeFind n given sum condition
DifficultyStandard +0.3 This is a straightforward arithmetic progression problem requiring identification of first term and common difference, application of the sum formula S_n = n/2[2a + (n-1)d], and solving a quadratic equation. While it involves multiple steps, each is routine and the problem type is standard textbook fare, making it slightly easier than average.
Spec1.04h Arithmetic sequences: nth term and sum formulae
1 The first term of an arithmetic progression is 61 and the second term is 57. The sum of the first \(n\) terms in \(n\). Find the value of the positive integer \(n\).
AnswerMarks Guidance
\(\frac{n}{2}[122 + (n-1)(-4)]\)M1 Attempt sum formula with \(a = 61, d = -4\)
\(n = \frac{n}{2}[122 + (n-1)(-4)]\)A1 Equated to \(n\) cao
\(2n(n-31) = 0\)DM1 Attempt to solve. Accept div. by \(n\) cao
\(n = 31\)A1 [4] 
$\frac{n}{2}[122 + (n-1)(-4)]$ | M1 | Attempt sum formula with $a = 61, d = -4$
$n = \frac{n}{2}[122 + (n-1)(-4)]$ | A1 | Equated to $n$ cao
$2n(n-31) = 0$ | DM1 | Attempt to solve. Accept div. by $n$ cao
$n = 31$ | A1 | [4]
1 The first term of an arithmetic progression is 61 and the second term is 57. The sum of the first $n$ terms in $n$. Find the value of the positive integer $n$.

\hfill \mbox{\textit{CAIE P1 2012 Q1 [4]}}
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