Sequences, nth term & arithmetic series – IGCSE Maths exam questions with mark scheme

1Grade 4–5HigherCalculator4 marks

Here are the first four terms of an arithmetic series.

$5 + 9 + 13 + 17 + \ldots$

(a) Find the 20th term of the series. [2]

(b) Find the sum of the first 20 terms of the series. [2]

(a) for $5 + (20 - 1) \times 4$ or $4n + 1$ with $n = 20$M1
$81$A1
(b) for $\dfrac{20}{2}(2 \times 5 + 19 \times 4)$ or $\dfrac{20}{2}(5 + 81)$ (ft their 81)M1
$860$A1
Answer: (a) 81 (b) 860
Examiner tip: Use (n − 1)d, not nd: the first term already counts as one of the 20 terms.

2Grade 6–7HigherCalculator5 marks

The arithmetic series $3 + 7 + 11 + 15 + \ldots$ has first term 3.

(a) Show that the sum of the first $n$ terms of the series is $2n^2 + n$ [2]

The sum of the first $k$ terms of the series is 595

(b) Find the value of $k$. [3]

(a) for $\dfrac{n}{2}(2 \times 3 + (n - 1) \times 4)$M1
for $\dfrac{n}{2}(4n + 2) = 2n^2 + n$ with all steps shown (shown)A1
(b) for $2k^2 + k = 595$ or $2k^2 + k - 595 = 0$M1
for a correct method to solve, e.g. $(2k + 35)(k - 17) = 0$ or correct substitution into the formulaM1
$k = 17$ (and $-17.5$ rejected or not given)A1
Answer: (b) $k = 17$
Examiner tip: For a 'show that' every line of algebra must appear — jumping straight to 2n² + n earns no A mark.

3Grade 6–7HigherCalculator5 marks

The 4th term of an arithmetic series is 23

The 10th term of the series is 5

(a) Find the first term and the common difference of the series. [3]

(b) Find the sum of the first 15 terms of the series. [2]

(a) for $a + 3d = 23$ and $a + 9d = 5$, or $6d = 5 - 23$M1
for a correct method to solve, e.g. $d = -3$ substituted to find $a$M1
$a = 32$, $d = -3$A1
(b) for $\dfrac{15}{2}(2 \times 32 + 14 \times (-3))$ (ft their $a$ and $d$)M1
$165$A1
Answer: (a) first term 32, common difference −3 (b) 165
Examiner tip: The gap from the 4th to the 10th term is 6 differences, not 7 — count the steps, not the terms.

4Grade 6–7HigherCalculator3 marks

Work out the sum of all the multiples of 7 between 1 and 300 [3]

for identifying the last term as $294$ or the number of terms as $42$M1
for $\dfrac{42}{2}(7 + 294)$ or $\dfrac{42}{2}(2 \times 7 + 41 \times 7)$ (ft their number of terms)M1
$6321$A1
Answer: 6321
Examiner tip: 300 ÷ 7 = 42.86…, so there are 42 multiples — rounding up to 43 includes 301, which is outside the range.

5Grade 6–7HigherCalculator5 marks

Maria saves money each week for 52 weeks.

She saves £20 in week 1. Each week after that, she saves £$x$ more than she saved the week before.

In total, she saves £3692 in the 52 weeks.

(a) Work out the value of $x$. [3]

(b) Find the first week in which Maria saves more than £100 [2]

(a) for $\dfrac{52}{2}(2 \times 20 + 51x) = 3692$ oeM1
for a correct first step to isolate $x$, e.g. $40 + 51x = 142$ or $1040 + 1326x = 3692$M1
$x = 2$A1
(b) for $20 + 2(n - 1) > 100$ oe, or finding that week 41 is exactly £100 (ft their $x$)M1
week $42$A1
Answer: (a) $x = 2$ (b) week 42
Examiner tip: In week 41 she saves exactly £100, which is not MORE than £100 — read inequalities in the question carefully.

6Grade 8–9HigherCalculator5 marks

An arithmetic series has first term 7 and common difference $d$.

The sum of the first 10 terms of the series is four times the sum of the first 4 terms.

Find the 20th term of the series. [5]

for $S_{10} = \dfrac{10}{2}(2 \times 7 + 9d)$ or $70 + 45d$M1
for $S_4 = \dfrac{4}{2}(2 \times 7 + 3d)$ or $28 + 6d$M1
for $70 + 45d = 4(28 + 6d)$ oe and simplifying to $21d = 42$ oeM1
$d = 2$A1
$45$ (ft $7 + 19 \times$ their $d$)B1
Answer: 45
Examiner tip: Put the 4 outside a bracket around the whole of S₄ — multiplying only the first term is the most common slip.

7Grade 4–5Calculator2 marks

Here are the first three terms of an arithmetic sequence.

$4 \qquad 11 \qquad 18$

Is 150 a term of this sequence? You must show how you get your answer. [2]

for the $n$th term $7n - 3$ (or $7n - 3 = 150$), or for listing/finding consecutive terms either side of 150, e.g. $144$ and $151$; allow $(150 - 4) \div 7$ oeM1
No, with a correct reason, e.g. $n = \frac{153}{7} = 21.857\ldots$ is not a whole number, or $144$ and $151$ are consecutive terms so 150 is skipped, or $(150-4) \div 7 = 20.857\ldots$ is not a whole number. A bare 'No' scores 0.A1
Answer: No — $7n - 3 = 150$ gives $n = \frac{153}{7}$, which is not an integer
Examiner tip: A bare 'no' scores nothing; you must show that n is not a whole number.

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