What is 4.4: Arithmetic sequences and series?

An arithmetic sequence has constant difference d and nth term u_n=a+(n-1)d, where a is the first term.

How many marks are 4.4 questions worth?

4.4 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 4.4 appeared in?

We have not yet indexed a verified past-paper appearance for 4.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

4.4

Understand and work with arithmetic sequences and series, including the formulae for nth term and the sum to n terms.

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Arithmetic sequences and series

Worked answers and methods for 4.4 on Edexcel A-level Maths 9MA0.

Explanation

An arithmetic sequence has constant difference $d$ and nth term $u_n=a+(n-1)d$ , where $a$ is the first term.
The sum of its first $n$ terms is $S_n=n[2a+(n-1)d]/2$ , equivalently $S_n=n(a+l)/2$ when the last term $l$ is known.
Pairing the first and last terms, then the second and penultimate terms, gives $n$ equal pairs of total $a+l$ across two copies of the series, proving $S_n=n(a+l)/2$ .
The term number $n$ must be a positive integer.
A common error is to use $a+nd$ for the nth term, shifting every term by one difference.

Worked example

An arithmetic series has first term $7$ , last term $151$ and common difference $8$ . Find the number of terms and the sum of the series.

1.Use $151=7+(n-1)8$ , giving $144=8(n-1)$ and $n=19$ .
2.Then $S_{19}=19(7+151)/2=19\times79=1501$ .

Answer: $19$ terms; $1501$

Common mistakes

Don't solve $u_n=L$ and round a non-integer term number instead of checking whether $L$ is actually a term.
Don't use the last term as the number of terms in the arithmetic-series formula.

Exam tip

Find the integer term count from the nth-term formula before applying the arithmetic sum formula.

Worked practice

Tier 1 · Easy

Find the

18

th term of the arithmetic sequence

11,17,23,\ldots

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$113$	2
Notes
Here $a=11$ and $d=6$ . Thus $u_{18}=11+(18-1)6=11+102=113$ .

(2 marks)

Tier 2 · Standard

The fifth term of an arithmetic sequence is

18

and the twelfth term is

46

. Find the first term, the common difference and the sum of the first

20

terms.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	First term $2$ Common difference $4$ $S_{20}=800$	5
Notes
If the first term is $a$ and common difference is $d$ , then $a+4d=18$ and $a+11d=46$ . Subtracting gives $7d=28$ , so $d=4$ and then $a=2$ . Hence $S_{20}=\dfrac{20}{2}[2(2)+19(4)]=10(80)=800$ .

(5 marks)

Tier 3 · Hard

An arithmetic sequence has first term

a

and common difference

d

. Its

8

th term is

31

and the sum of its first

20

terms is

730

. Find

a

and

d

, and find the least value of

n

for which the nth term exceeds

100

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$a=\dfrac{78}{5}$ $d=\dfrac{11}{5}$ $n=40$	5
Notes
The term condition gives $a+7d=31$ . The sum condition gives $10(2a+19d)=730$ , so $2a+19d=73$ . Doubling the first equation and subtracting gives $5d=11$ , hence $d=11/5$ and $a=31-77/5=78/5$ . Therefore $u_n=(11n+67)/5$ . Solving $(11n+67)/5>100$ gives $n>433/11=39.36\ldots$ , so the least integer is $40$ .

(5 marks)

Tier 1 · Easy

The seventh term of an arithmetic sequence is

29

and the common difference is

4

. Find the first term and the fifteenth term.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	First term $5$ Fifteenth term $61$	3
Notes
If the first term is $a$ , then $a+6(4)=29$ , so $a=5$ . The fifteenth term is $a+14d=5+14(4)=61$ .

(3 marks)

Tier 2 · Standard

An arithmetic sequence has partial sum

S_n=3n^2+2n

. Derive its

n

th term and hence determine the position of the term equal to

167

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$u_n=6n-1$ The $28$ th term	4
Notes
For $n\geq2$ , $u_n=S_n-S_{n-1}$ . Since $S_{n-1}=3(n-1)^2+2(n-1)=3n^2-4n+1$ , it follows that $u_n=6n-1$ ; this also gives $u_1=5=S_1$ . Solving $6n-1=167$ gives $n=28$ .

(4 marks)

Tier 3 · Hard

An arithmetic sequence has positive first term

a

and positive common difference

d

. Its first, fifth and ninth terms are the side lengths, in increasing order, of a right-angled triangle with perimeter

72

. Find

a

and

d

, and then calculate

S_{20}

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$a=18$ $d=\dfrac32$ $S_{20}=645$	6
Notes
The three lengths are $a$ , $a+4d$ and $a+8d$ , with the last as hypotenuse. Pythagoras gives $a^2+(a+4d)^2=(a+8d)^2$ , so $a^2-8ad-48d^2=0$ , or $(a-12d)(a+4d)=0$ . Positivity gives $a=12d$ . The perimeter condition is $3a+12d=72$ , hence $48d=72$ , so $d=3/2$ and $a=18$ . Therefore $S_{20}=10[2(18)+19(3/2)]=645$ .

(6 marks)

Tier 2 · Standard

Five numbers are inserted between

7

and

49

so that all seven numbers form an arithmetic sequence. Find the five inserted numbers and their sum.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$14,21,28,35,42$ Sum $=140$	3
Notes
Across the seven terms there are six equal steps, so the common difference is $(49-7)/6=7$ . The inserted terms are therefore $14,21,28,35,42$ , and their sum is $140$ .

(3 marks)

Tier 3 · Hard

An arithmetic sequence has first term

30

and common difference

-4

. Let

S_n

denote the sum of its first

n

terms. Determine the positive integer

n

for which

S_n

is greatest and find this greatest value. Find also all positive integers

n

for which

S_n\geq96

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$S_n$ is greatest when $n=8$ Greatest value $=128$ $4\leq n\leq12$ , $n$ an integer	5
Notes
$S_n=\dfrac n2[60-4(n-1)]=2n(16-n)=128-2(n-8)^2$ . Thus the greatest value is $128$ at the positive integer $n=8$ . Also $S_n\geq96$ gives $2n(16-n)\geq96$ , or $(n-4)(n-12)\leq0$ . Hence the positive integers are $4\leq n\leq12$ .

(5 marks)

Tier 3 · Hard

The arithmetic sequences

7,12,17,\ldots

and

4,11,18,\ldots

have some terms in common. Find the first four common terms. Hence find the sum of all their common terms that are less than

500

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$32,67,102,137$ Sum $=3633$	6
Notes
Listing terms until the first match gives $7,12,17,22,27,32$ and $4,11,18,25,32$ , so the first common term is $32$ . The gap between successive common terms must be a multiple of both $5$ and $7$ . The positive multiples of $5$ below $35$ are $5,10,15,20,25,30$ , and none is a multiple of $7$ , while $35=5\times7$ . Hence the smallest possible positive gap is $35$ , and adding $35$ always produces another term of each sequence. The first four common terms are therefore $32,67,102,137$ . Those below $500$ are $32+35j$ for $j=0,1,\ldots,13$ , since the next term is $522$ . Their sum is $14[2(32)+13(35)]/2=3633$ .

(6 marks)

Q10

Tier 3 · Hard

10.

The digits of a three-digit positive integer form an increasing arithmetic sequence. The integer is

15

times the sum of its digits. Find the integer and prove that it is the only possibility.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$135$	5
Notes
Let the hundreds digit be $a$ and the positive common difference be $d$ . The digits are $a,a+d,a+2d$ , so the integer is $100a+10(a+d)+(a+2d)=111a+12d$ . The digit sum is $3a+3d$ . The given condition gives $111a+12d=45a+45d$ , hence $d=2a$ . The units digit is then $a+2d=5a$ . Since $a$ is a positive digit and $5a\leq9$ , the only possibility is $a=1$ . Thus $d=2$ and the integer is $135$ , which indeed equals $15(1+3+5)$ .

(5 marks)

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Other points in 4 Sequences and series

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