Question 4 - A-Level Maths

CAIE P1 2020 November — Question 4 5 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2020
Session	November
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Arithmetic Sequences and Series
Type	Find n given sum condition
Difficulty	Moderate -0.3 This is a straightforward application of standard formulas: finding the nth term from the sum formula (using u_n = S_n - S_{n-1} gives u_n = 2n + 3), then solving a simple linear inequality 2k + 3 > 200 to get k > 98.5, so k = 99. Slightly easier than average as it requires only routine algebraic manipulation with no conceptual challenges.
Spec	1.04h Arithmetic sequences: nth term and sum formulae

4 The sum, $S _ { n }$, of the first $n$ terms of an arithmetic progression is given by $$S _ { n } = n ^ { 2 } + 4 n$$ The $k$ th term in the progression is greater than 200.
Find the smallest possible value of $k$.

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Question 4:

Answer	Marks	Guidance
Answer	Mark	Guidance
$S_x$ and $S_{x+1}$	M1	Using two values of $n$ in the given formula
$a = 5,\ d = 2$	A1 A1
$a + (n-1)d > 200 \rightarrow 5 + 2(k-1) > 200$	M1	Correct formula used with their $a$ and $d$ to form an equation or inequality with 200, condone use of $n$
$(k =)\ 99$	A1	Condone $\geqslant 99$
Alternative: $\frac{n}{2}(2a + (n-1)d) \equiv n^2 + 4n \rightarrow \left(\frac{d}{2} = 1,\ a - \frac{1}{2}d = 4\right)$	M1	Equating two correct expressions of $S_n$ and equating coefficients of $n$ and $n^2$
$d = 2,\ a = 5$	A1 A1
$a + (n-1)d > 200 \rightarrow 5 + 2(k-1) > 200$	M1	Correct formula used with their $a$ and $d$
$(k =)\ 99$	A1	Condone $\geqslant 99$
Alternative: $\text{sum}_k - \text{sum}_{k-1} \rightarrow k^2 + 4k - (k-1)^2 - 4(k-1)$	M1 A1	Using given formula with consecutive expressions subtracted. Allow $k+1$ and $k$.
$2k + 3 > 200$ or $= 200$	M1 A1	Simplifying to a linear equation or inequality
$(k =)\ 99$	A1	Condone $\geqslant 99$

## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| $S_x$ and $S_{x+1}$ | M1 | Using two values of $n$ in the given formula |
| $a = 5,\ d = 2$ | A1 A1 | |
| $a + (n-1)d > 200 \rightarrow 5 + 2(k-1) > 200$ | M1 | Correct formula used with their $a$ and $d$ to form an equation or inequality with 200, condone use of $n$ |
| $(k =)\ 99$ | A1 | Condone $\geqslant 99$ |
| **Alternative:** $\frac{n}{2}(2a + (n-1)d) \equiv n^2 + 4n \rightarrow \left(\frac{d}{2} = 1,\ a - \frac{1}{2}d = 4\right)$ | M1 | Equating two correct expressions of $S_n$ and equating coefficients of $n$ and $n^2$ |
| $d = 2,\ a = 5$ | A1 A1 | |
| $a + (n-1)d > 200 \rightarrow 5 + 2(k-1) > 200$ | M1 | Correct formula used with their $a$ and $d$ |
| $(k =)\ 99$ | A1 | Condone $\geqslant 99$ |
| **Alternative:** $\text{sum}_k - \text{sum}_{k-1} \rightarrow k^2 + 4k - (k-1)^2 - 4(k-1)$ | M1 A1 | Using given formula with consecutive expressions subtracted. Allow $k+1$ and $k$. |
| $2k + 3 > 200$ or $= 200$ | M1 A1 | Simplifying to a linear equation or inequality |
| $(k =)\ 99$ | A1 | Condone $\geqslant 99$ |

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Show LaTeX source

4 The sum, $S _ { n }$, of the first $n$ terms of an arithmetic progression is given by

$$S _ { n } = n ^ { 2 } + 4 n$$

The $k$ th term in the progression is greater than 200.\\
Find the smallest possible value of $k$.\\

\hfill \mbox{\textit{CAIE P1 2020 Q4 [5]}}

This paper (11 questions)

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Q1 4 Q2 5 Q3 5 Q4 5 Q5 6 Q6 6 Q7 7 Q8 7 Q9 9 Q10 10 Q11 11

Answer	Marks	Guidance
Answer	Mark	Guidance
\(S_x\) and \(S_{x+1}\)	M1	Using two values of \(n\) in the given formula
\(a = 5,\ d = 2\)	A1 A1
\(a + (n-1)d > 200 \rightarrow 5 + 2(k-1) > 200\)	M1	Correct formula used with their \(a\) and \(d\) to form an equation or inequality with 200, condone use of \(n\)
\((k =)\ 99\)	A1	Condone \(\geqslant 99\)
Alternative: \(\frac{n}{2}(2a + (n-1)d) \equiv n^2 + 4n \rightarrow \left(\frac{d}{2} = 1,\ a - \frac{1}{2}d = 4\right)\)	M1	Equating two correct expressions of \(S_n\) and equating coefficients of \(n\) and \(n^2\)
\(d = 2,\ a = 5\)	A1 A1
\(a + (n-1)d > 200 \rightarrow 5 + 2(k-1) > 200\)	M1	Correct formula used with their \(a\) and \(d\)
\((k =)\ 99\)	A1	Condone \(\geqslant 99\)
Alternative: \(\text{sum}_k - \text{sum}_{k-1} \rightarrow k^2 + 4k - (k-1)^2 - 4(k-1)\)	M1 A1	Using given formula with consecutive expressions subtracted. Allow \(k+1\) and \(k\).
\(2k + 3 > 200\) or \(= 200\)	M1 A1	Simplifying to a linear equation or inequality
\((k =)\ 99\)	A1	Condone \(\geqslant 99\)