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Sec 4 E Maths: Number Patterns and Sequences, practice questions & worked solutions

Number patterns questions selected from 2023 to 2025 Singapore school E Maths prelim papers, each with the school’s full worked solution.

8 questionsStep-by-step solutionsFree to access

About this topic & key methods

Number patterns start in Sec 1 with linear sequences, and E Maths prelims extend them to quadratic rules, written patterns and simple proofs about terms. Most questions here are Paper 1, with one longer Paper 2 question.

Attempt each question on paper before opening its worked solution. The solutions are the schools’ own marking schemes, so some steps are brief.

Key methods for number patterns and sequences

For a linear sequence, the nth term is $a+(n-1)d$ , where $a$ is the first term and $d$ the common difference
To test whether $N$ is a term, solve $T_n=N$ and check that $n$ is a positive integer
Compare a new sequence with a known one term by term, for example each term is 7 less than $2n^2+4$
Write line $n$ of a pattern by tracking what changes with $n$ , then simplify
Use $T_{n+1}-T_n$ to study how consecutive terms differ
In a sum of differences such as $(4^2-1^2)+(7^2-4^2)+\ldots$ , cancel the middle terms

For lessons that build these methods week by week, see our E Maths tuition for Sec 3 and Sec 4.

Number Patterns and Sequences questions with worked solutions

Unless the question specifies otherwise, give non-exact answers to 3 significant figures and angles in degrees to 1 decimal place.

Q1Boon Lay Secondary School2025Prelim P1 Q16●○○ Foundational5 marks

Linear sequence and testing whether a number is a term, Boon Lay Secondary School, 2025 Prelim P1 Q16

In a sequence, the same number is added each time to obtain the next number.
The first 4 terms of the sequence are

$5\quad a\quad b\quad26$

Find the values of $a$ and $b$ . [2]
Find an expression, in terms of $n$ , for the $n$ th term of the sequence. [2]
Explain whether 2025 is a term of this sequence. [1]

Show worked solution▾

(a)

$\frac{26-5}{3}=7$

$a=5+7=12$

$b=12+7=19$

(b)

$7n-2$

(c)

$7n-2=2025$

$n=289\frac47$

which is not an integer, hence not a term.

Answer: $\text{(a) }a=12,\ b=19\quad\text{(b) }7n-2\quad\text{(c) No}$

Q2Yio Chu Kang Secondary School2025Prelim P1 Q7●●○ Standard2 marks

Finding missing terms of a linear sequence from their sum, Yio Chu Kang Secondary School, 2025 Prelim P1 Q7

In this sequence, the difference between any two consecutive terms is the same number.

$w\quad x\quad y\quad23\quad z\quad\ldots$

The sum of the first five terms is 80. Find the values of $w$ , $x$ , $y$ and $z$ . [2]

Show worked solution▾

$\begin{aligned}(23-3d)+(23-2d)+(23-d)+23+(23+d)&=80\\-5d+115&=80\\d&=7\end{aligned}$

$w=2,\ x=9,\ y=16,\ z=30$

Answer: $w=2,\ x=9,\ y=16,\ z=30$

Q3Tampines Secondary School2024Prelim P1 Q20●●○ Standard3 marks

Quadratic nth term by comparing two sequences, Tampines Secondary School, 2024 Prelim P1 Q20

The $n$ th term of a sequence is given by $2n^2+4$ .
1. Write down the first 5 terms. [1]
2. Explain why it is not possible for a term in this sequence to be an odd number. [1]
The first 5 terms of another sequence are $-1$ , $5$ , $15$ , $29$ , $47$ , $\ldots$
By comparing this sequence with your answer to (a), write down the $k$ th term. [1]

Show worked solution▾

(a)(i)

6, 12, 22, 36, 54

(a)(ii)

$2n^2+4=2(n^2+2)$ is a multiple of 2 for all values of $n$ , hence it is an even number and not an odd number.

(b)

$2k^2-3$

Answer: (a)(i) 6, 12, 22, 36, 54 (ii) $2n^2+4=2(n^2+2)$ is a multiple of 2, so no term is odd (b) $2k^2-3$

Q4Ahmad Ibrahim Secondary School2023Prelim P1 Q20●●○ Standard4 marks

Difference of two squares in a number pattern, Ahmad Ibrahim Secondary School, 2023 Prelim P1 Q20

The first four terms in a sequence of numbers are given below.

$\begin{aligned}T_1&=3^2-(-1)^2=8\\T_2&=4^2-0^2=16\\T_3&=5^2-1^2=24\\T_4&=6^2-2^2=32\end{aligned}$

Write down an expression, in the same form, in terms of $n$ , to represent $T_n$ . [1]
Hence find the positive integer value of $h$ and $k$ such that $h^2-k^2=600$ . [3]

Show worked solution▾

(a)

$T_n=(n+2)^2-(n-2)^2=8n$

(b)

$\begin{aligned}600\div8&=75\end{aligned}$

Pattern number is 75.

$\begin{aligned}h&=75+2=77\\k&=75-2=73\end{aligned}$

Answer: (a) $T_n=(n+2)^2-(n-2)^2=8n$ (b) $h=77$ , $k=73$

Q5CHIJ St. Nicholas Girls’ School2023Prelim P1 Q18●●○ Standard6 marks

Dot diagram sequence with a quadratic rule, CHIJ St. Nicholas Girls’ School, 2023 Prelim P1 Q18

The first four diagrams in a sequence are shown below.

CHIJ St. Nicholas Girls’ School 2023 P1 Q18: diagram for the question

Find the number of dots in Diagram 5. [1]
Find an expression, in terms of $n$ , for the number of dots in Diagram $n$ . [2]
Find the difference in the number of dots between Diagram 7 and Diagram 6. [1]
Show that the difference in the number of dots between Diagram $(n+1)$ and Diagram $n$ is $4n+3$ . [2]

Show worked solution▾

(a)

$5\times11=55$

(b)

$n(2n+1)$

(c)

Diagram 7 has 105 dots

Diagram 6 has 78 dots

difference $=105-78=27$

(d)

Diagram $(n+1)$ has $(n+1)(2n+3)$

$\begin{aligned}&(n+1)(2n+3)-n(2n+1)\\&=2n^2+5n+3-2n^2-n\\&=4n+3\end{aligned}$

Answer: (a) 55 (b) $n(2n+1)$ (c) 27

Q6Cedar Girls’ Secondary School2025Prelim P1 Q13●●● Challenging5 marks

Telescoping sum of a difference of squares pattern, Cedar Girls’ Secondary School, 2025 Prelim P1 Q13

Consider the pattern

$\begin{aligned}L_1&:4^2-1^2=15\\L_2&:5^2-2^2=21\\L_3&:6^2-3^2=27\\L_4&:7^2-4^2=33\\&\quad \vdots\\L_n&:\ldots\end{aligned}$

Write down $L_n$ of the pattern. [1]
Explain, with working, whether 3336 is a term of the sequence $15,21,27,33,\ldots$ . [2]
Find the exact value of $L_1+L_4+L_7+L_{10}+L_{13}+\ldots L_{97}+L_{100}$ . [2]

Show worked solution▾

(a)

$L_n=(n+3)^2-n^2=6n+9$

(b)

$3336=6n+9$

$n=554.5$

No, as $n$ is not a (positive) integer / whole number.

(c)

$\begin{aligned}&L_1+L_4+L_7+L_{10}+L_{13}+\ldots L_{97}+L_{100}\\&=4^2-1^2+7^2-4^2+10^2-7^2+\ldots+100^2-97^2+103^2-100^2\\&=103^2-1^2\\&=10608\end{aligned}$

Answer: $\text{(a) }L_n=(n+3)^2-n^2=6n+9\quad\text{(b) No}\quad\text{(c) }10608$

Q7Nan Chiau High School2024Prelim P1 Q21●●● Challenging6 marks

Generalising a pattern of squares and solving for the line, Nan Chiau High School, 2024 Prelim P1 Q21

Consider the following pattern.

Line 1:	$2^2+0^2=(2-0)^2+2(2)(0)$
Line 2:	$4^2+1^2=(4-1)^2+2(4)(1)$
Line 3:	$6^2+2^2=(6-2)^2+2(6)(2)$
Line 4:	$8^2+3^2=(8-3)^2+2(8)(3)$
	$\vdots$ $\vdots$
	$\vdots$ $\vdots$

Write down Line 5 of the sequence of pattern. [1]
Write down Line $n$ of the sequence of pattern. [1]
Hence find the value of $p$ and of $q$ , given that $p^2+q^2=1249$ is part of the pattern where $q<p$ . [4]

Show worked solution▾

(a)

$10^2+4^2=(10-4)^2+2(10)(4)$

(b)

$(2n)^2+(n-1)^2=(2n-n+1)^2+2(2n)(n-1)$
or $=4n^2+n^2-2n+1$ or $=5n^2-2n+1$

(c)

$\begin{aligned}p^2+q^2&=1249\\5n^2-2n+1&=1249\\5n^2-2n-1248&=0\\(n-16)(5n+78)&=0\\n=16\quad\text{or}\quad n&=-\frac{78}5\ \text{(rej)}\end{aligned}$

Hence, $p=32$ and $q=15$

Answer: (a) $10^2+4^2=(10-4)^2+2(10)(4)$ (b) $(2n)^2+(n-1)^2=(2n-n+1)^2+2(2n)(n-1)$ (c) $p=32$ and $q=15$

Q8Canberra Secondary School2024Prelim P2 Q8●●● Challenging8 marks

Sum of consecutive terms of a quadratic sequence, Canberra Secondary School, 2024 Prelim P2 Q8

The first four terms in a sequence of numbers are given below.

$\begin{aligned}T_1&=(1\times3)+8=11\\T_2&=(2\times4)+12=20\\T_3&=(3\times5)+16=31\\T_4&=(4\times6)+20=44\end{aligned}$

Find $T_5$ . [1]
Show that the $n^{\text{th}}$ term of the sequence, $T_n$ is given by $n^2+6n+4$ . [2]
$T_{p-1}$ and $T_p$ are consecutive terms in the sequence.
Find and simplify an expression, in terms of $p$ , for the sum of $T_{p-1}$ and $T_p$ . [3]
Find the value of $p$ when the sum of $T_{p-1}$ and $T_p$ is 303. [2]