OCR C2 2007 January — Question 1 4 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2007
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeSum of first n terms
DifficultyModerate -0.8 This is a straightforward application of standard arithmetic progression formulas requiring only three steps: find the common difference from a + 19d = 72, then apply S_n = n/2(2a + (n-1)d) with n=100. It's more routine than average since it involves direct formula substitution with no problem-solving insight needed, though the arithmetic with larger numbers prevents it from being trivial.
Spec1.04h Arithmetic sequences: nth term and sum formulae

In an arithmetic progression the first term is 15 and the twentieth term is 72. Find the sum of the first 100 terms. [4]

AnswerMarks Guidance
\(15 + 19d = 72\) Hence \(d = 3\) \(S_n = \frac{100}{2}[2(2×15) + (99×3)] = 16350\)M1, A1, M1, A1 (4 marks) Attempt to find \(d\), from \(a + (n-1)d\) or \(a + nd\); Obtain \(d = 3\); Use correct formula for sum of \(n\) terms; Obtain 16350
$15 + 19d = 72$ Hence $d = 3$ $S_n = \frac{100}{2}[2(2×15) + (99×3)] = 16350$ | M1, A1, M1, A1 (4 marks) | Attempt to find $d$, from $a + (n-1)d$ or $a + nd$; Obtain $d = 3$; Use correct formula for sum of $n$ terms; Obtain 16350
In an arithmetic progression the first term is 15 and the twentieth term is 72. Find the sum of the first 100 terms. [4]

\hfill \mbox{\textit{OCR C2 2007 Q1 [4]}}