CAIE P1 2017 November — Question 1 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2017
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicArithmetic Sequences and Series
TypeFind n given sum condition
DifficultyStandard +0.3 This is a straightforward application of the arithmetic series sum formula S_n = n/2[2a + (n-1)d] leading to a quadratic inequality. While it requires solving n/2[-24 + 6(n-1)] > 3000 and interpreting the result correctly (taking the ceiling), it's a standard textbook exercise with no conceptual challenges beyond routine algebraic manipulation.
Spec1.02g Inequalities: linear and quadratic in single variable1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum
1 An arithmetic progression has first term - 12 and common difference 6 . The sum of the first \(n\) terms exceeds 3000 . Calculate the least possible value of \(n\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{2}n[-24+(n-1)6] \sim 3000\)M1 Use correct formula with RHS \(\approx 3000\) (e.g. 3010). Note: ~ denotes any inequality or equality
\((3)(n^2-5n-1000)(\sim 0)\)A1 Rearrange into a 3-term quadratic
\(n \sim 34.2\) (& \(-29.2\))A1
35. Allow \(n \geqslant 35\)A1 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}n[-24+(n-1)6] \sim 3000$ | M1 | Use correct formula with RHS $\approx 3000$ (e.g. 3010). Note: ~ denotes any inequality or equality |
| $(3)(n^2-5n-1000)(\sim 0)$ | A1 | Rearrange into a 3-term quadratic |
| $n \sim 34.2$ (& $-29.2$) | A1 | |
| 35. Allow $n \geqslant 35$ | A1 | |

---
1 An arithmetic progression has first term - 12 and common difference 6 . The sum of the first $n$ terms exceeds 3000 . Calculate the least possible value of $n$.\\

\hfill \mbox{\textit{CAIE P1 2017 Q1 [4]}}
This paper (11 questions)
View full paper
Q1 4 Q2 4 Q3 5 Q4 5 Q5 7 Q6 7 Q7 7 Q8 8 Q9 9 Q10 9 Q11 10