MathsICSE Class 10 · Arithmetic and Geometric Progression

nth Term of an A.P. (General Term): Formula and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

In an arithmetic progression with first term a and common difference d, the nth term is Tn = a + (n − 1)d. Use it to find any term, to find which term equals a given number (solve for n, which must be a positive integer), or to rebuild the A.P. from two known terms.

To reach the nth term from a you climb only n − 1 steps of size d, so the 5th term is a + 4d, not a + 5d.

Illustration drawn for QuestionBuddy.

Understand nth Term of an A.P.

An arithmetic progression (A.P.) is a list of numbers in which each term is obtained from the one before by adding the same fixed number, the common difference $d = T_{2} - T_{1} = T_{3} - T_{2} = \dots$ . The difference can be positive, negative, zero or a fraction.

The general term. Starting from $a$ , the 2nd term adds $d$ once, the 3rd adds it twice, and the $n$ th adds it $(n - 1)$ times:

$T_{n} = a + (n - 1) d .$

The examiners reported in 2024 Q1(viii) that candidates added $7 d$ for the 7th term instead of $6 d$ . Remember: $n - 1$ jumps, not $n$ .

Four question types.

1. Find a term. Read $a$ and $d$ from the list and substitute.
2. Which term is a given number? Set $a + (n - 1) d$ equal to it and solve for $n$ . If $n$ is not a positive whole number, the number is not a term.
3. Two terms given. For example $T_{4} = 11$ and $T_{10} = 29$ give $a + 3 d = 11$ and $a + 9 d = 29$ ; subtract to get $6 d = 18$ . In general, $T_{q} - T_{p} = (q - p) d$ .
4. ** $T_{n}$ given as a formula.** If $T_{n} = 6 (7 - n)$ (2024 Q6(ii)), put $n = 1$ and $n = 2$ : $T_{1} = 36$ , $T_{2} = 30$ , so $a = 36$ , $d = - 6$ . A formula that is linear in $n$ , $T_{n} = p n + q$ , always gives an A.P. with $d = p$ .

Counting from the end. For an A.P. ending at last term $l$ , the $n$ th term from the end is $l - (n - 1) d$ . The number of terms in $a, a + d, \dots, l$ is $\frac{l - a}{d} + 1$ .

Three terms in A.P. $p$ , $q$ , $r$ are in A.P. exactly when $q - p = r - q$ , that is $2 q = p + r$ (the middle term is the average of its neighbours). When three unknown terms are needed, take them as $a - d$ , $a$ , $a + d$ .

nth Term of an A.P.: formulas to remember

Common difference: $d = T_{2} - T_{1} = T_{n} - T_{n - 1}$; Later term minus earlier term.
nth term: $T_{n} = a + (n - 1) d$; $n - 1$ jumps of size $d$ .
Difference of two terms: $T_{q} - T_{p} = (q - p) d$; Quick route when two terms are given.
nth term from the end: $l - (n - 1) d$; $l$ is the last term.
Number of terms: $n = \frac{l - a}{d} + 1$; Counting terms from $a$ to $l$ .
Three terms in A.P.: $2 q = p + r$; Middle term is the average of the other two.

How to solve nth Term of an A.P. questions

Identify a and d, write the general term, then use it as an equation.

1Write $a = T_{1}$ and $d = T_{2} - T_{1}$ (check with $T_{3} - T_{2}$ ).
2Write $T_{n} = a + (n - 1) d$ with the numbers substituted and simplify, for example $T_{n} = 3 n - 1$ .
3To find a term, put the value of $n$ in.
4To test a number, set $T_{n}$ equal to it and solve; accept only a positive integer $n$ .
5With two terms given, write two equations in $a$ and $d$ and subtract.
6State the answer in words, for example 'the 35th term is −81'.

nth Term of an A.P. in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Using Tn to find when a target is reached

Suppose a sapling is 45 cm tall when planted and grows 6 cm every month, so its heights at the end of months 1, 2, 3, … are 51, 57, 63, … cm.

At the end of which month does it first reach 1.5 m?

Try it first, then show the solutionSolution

$a = 51$ , $d = 6$ : $T_{n} = 51 + (n - 1) 6 = 45 + 6 n$ .
$45 + 6 n \geq 150$ gives $n \geq 17.5$ .
$T_{17} = 147$ cm (not yet); $T_{18} = 153$ cm.

Answer: At the end of month 18 (height 153 cm).

Memory hook: If n comes out as a fraction, the target falls between two terms; round up for 'first reaches'.

A decreasing A.P.

Imagine a phone whose battery reads 92% at 9:00 and then 3% less at each reading taken every 15 minutes: 92, 89, 86, …

Which reading is the first below 20%, and at what time is it taken?

Try it first, then show the solutionSolution

$T_{n} = 92 + (n - 1) (- 3) = 95 - 3 n$ .
$T_{25} = 20$ (not below), $T_{26} = 17$ .
Reading 26 is $25 \times 15 = 375$ minutes after 9:00, that is 6 hours 15 minutes later.

Answer: The 26th reading, 17%, at 15:15.

Memory hook: Reading 1 is at time 0, so reading n is (n − 1) steps later.

Tricks to remember nth Term of an A.P.

Short memory aids. Say each one aloud once, then use it on a question the same day.

Fence posts
From post 1 to post $n$ there are $n - 1$ gaps; from $T_{1}$ to $T_{n}$ there are $n - 1$ jumps of $d$ .
Use it when: Avoiding $a + n d$ , the error reported in 2024 Q1(viii).
Two terms: subtract to kill a
$T_{10} - T_{4} = 6 d$ : the gap in positions times $d$ .
Use it when: A.P.s from two given terms.
Linear in n means A.P.
$T_{n} = p n + q$ is an A.P. with $d = p$ (the number in front of $n$ ).
Use it when: Questions that give $T_{n}$ as a formula.

nth Term of an A.P.: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A straightforward term · 1 mark · MCQBoard level

The 12th term of the A.P. 4, 9, 14, … is:
(a) 59
(b) 64
(c) 60
(d) 55

Try it first, then show the solutionSolution

$a = 4$ , $d = 5$ .
$T_{12} = 4 + 11 \times 5 = 59$ . Option (b) uses $12 \times 5$ by mistake.

Answer: (a) 59

Example 2: Which term is it? · 3 marks · Short answerBoard level

Which term of the A.P. 21, 18, 15, … is −81?

Try it first, then show the solutionSolution

$a = 21$ , $d = 18 - 21 = - 3$ .
$21 + (n - 1) (- 3) = - 81 \Rightarrow (n - 1) (- 3) = - 102 \Rightarrow n - 1 = 34$ .
$n = 35$ .

Answer: The 35th term.

Example 3: From two terms, and testing a number · 3 marks · Short answerHarder board level

The 4th term of an A.P. is 11 and the 10th term is 29.
(a) Find the first term and the common difference.
(b) Is 100 a term of this A.P.?

Try it first, then show the solutionSolution

(a) $a + 3 d = 11$ and $a + 9 d = 29$ . Subtracting: $6 d = 18$ , so $d = 3$ and $a = 2$ .
(b) $T_{n} = 2 + (n - 1) 3 = 3 n - 1$ . Setting $3 n - 1 = 100$ gives $n = \frac{101}{3}$ , not a whole number.

Answer: (a) $a = 2$ , $d = 3$ (b) No, 100 is not a term.

Example 4: Tn given as a formula · 4 marks · Short answerHarder board level

The $n$ th term of a sequence is $T_{n} = 18 - 4 n$ .
(a) Show that it is an A.P. and state $a$ and $d$ .
(b) Find the first negative term and its position.

Try it first, then show the solutionSolution

(a) $T_{n} - T_{n - 1} = (18 - 4 n) - (18 - 4 n + 4) = - 4$ , a constant, so it is an A.P. $T_{1} = 14$ , so $a = 14$ , $d = - 4$ .
(b) $18 - 4 n < 0 \Rightarrow n > 4.5$ , so the first negative term is $T_{5}$ .
$T_{5} = 18 - 20 = - 2$ (check: $T_{4} = 2$ is still positive).

Answer: (a) $a = 14$ , $d = - 4$ (b) The 5th term, −2.

Example 5: Seven times the 7th equals eleven times the 11th · 4 marks · Short answerHOTS / competency

In an A.P. with non-zero common difference, 7 times the 7th term equals 11 times the 11th term. Show that the 18th term is zero.

Try it first, then show the solutionSolution

$7 (a + 6 d) = 11 (a + 10 d)$ .
$7 a + 42 d = 11 a + 110 d \Rightarrow - 4 a = 68 d \Rightarrow a = - 17 d$ .
$T_{18} = a + 17 d = - 17 d + 17 d = 0$ .

Answer: $T_{18} = 0$ , as required.

Example 6: Case study: a taxi fare · 4 marks · Case-basedHOTS / competency

Suppose a taxi charges ₹50 for the first kilometre and ₹16 for each further kilometre.
(a) Show that the fares for 1, 2, 3, … km form an A.P.
(b) Find the fare for 12 km.
(c) How far can you travel for ₹386?

Try it first, then show the solutionSolution

(a) Fares 50, 66, 82, …: each is 16 more than the last, so an A.P. with $a = 50$ , $d = 16$ .
(b) $T_{12} = 50 + 11 \times 16 = 226$ .
(c) $50 + (n - 1) 16 = 386 \Rightarrow n - 1 = 21 \Rightarrow n = 22$ .

Answer: (b) ₹226 (c) 22 km

Where students lose marks in nth Term of an A.P.

The slips examiners see most often, and what to write instead.

Mistake: Using nd instead of (n − 1)d
There are $n - 1$ steps from the 1st term to the $n$ th.
Wrong working
$T_{7} = \frac{1}{a} + 7$
Right working
$T_{7} = \frac{1}{a} + 6 \times 1 = \frac{1}{a} + 6$
Do this: Write the formula $T_{n} = a + (n - 1) d$ before substituting.
Usually costs: The mark in 2024 Q1(viii), where the examiners reported this error.
Mistake: Not finding a and d from a formula for Tn
When $T_{n}$ is given, $a$ and $d$ come from $T_{1}$ and $T_{2} - T_{1}$ .
Wrong working
$T_{n} = 6 (7 - n)$ , so $a = 6$ and $d = 7$ .
Right working
$T_{1} = 36$ , $T_{2} = 30$ , so $a = 36$ , $d = - 6$ .
Do this: Substitute $n = 1$ and $n = 2$ every time.
Usually costs: Reported for 2024 Q6(ii); all marks for that part.
Mistake: Wrong sign of d
$d$ is later term minus earlier term; a falling A.P. has negative $d$ .
Wrong working
21, 18, 15: $d = 3$ .
Right working
$d = 18 - 21 = - 3$ .
Do this: Always compute $T_{2} - T_{1}$ , never 'the gap'.
Usually costs: Every later answer.

Exam tips

2024 Q1(viii) (MCQ) tested only $T_{7} = a + 6 d$ ; the trap option added 7.
2024 Q6(ii) gave $T_{n} = 6 (7 - n)$ ; the expected first step was $T_{1} = 36$ , $T_{2} = 30$ .
When you solve for $n$ , say clearly whether it is a natural number; a fraction means 'not a term'.
Number your two equations when finding $a$ and $d$ from two terms, and show the subtraction.

Practice questions on nth Term of an A.P.

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The common difference of the A.P. 7, 3, −1, … is:
(a) 4
(b) −4
(c) −3
(d) 3
Hint
Second term minus first.
Answer
(b) −4
Q2 · 2 marks · Short answerBoard level
Find the 15th term of the A.P. −5, −2, 1, …
Hint
$a = - 5$ , $d = 3$ .
Answer
$- 5 + 14 \times 3 = 37$
Q3 · 3 marks · Short answerHarder board level
Is 184 a term of the A.P. 3, 7, 11, …? Give a reason.
Hint
Solve $3 + 4 (n - 1) = 184$ .
Answer
$4 (n - 1) = 181$ gives $n = 46.25$ , not a whole number, so 184 is not a term.
Q4 · 3 marks · Short answerHarder board level
Find $k$ if $2 k - 1$ , $k + 3$ and $3 k + 1$ are consecutive terms of an A.P.
Hint
Use $2 \times middle = sum of the outer two$ .
Answer
$2 (k + 3) = 5 k$ gives $k = 2$ ; the terms are 3, 5, 7.
Q5 · 3 marks · Short answerHarder board level
How many two-digit numbers are divisible by 3?
Hint
The A.P. 12, 15, …, 99.
Answer
$\frac{99 - 12}{3} + 1 = 30$
Q6 · 3 marks · Short answerHarder board level
The 3rd term of an A.P. is 16 and the 7th term exceeds the 5th term by 12. Find the A.P.
Hint
$T_{7} - T_{5} = 2 d$ .
Answer
$2 d = 12$ , $d = 6$ ; $a + 2 d = 16$ gives $a = 4$ . The A.P. is 4, 10, 16, 22, …
Q7 · 4 marks · Short answerHOTS / competency
The sum of the 4th and 8th terms of an A.P. is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms.
Hint
$2 a + 10 d = 24$ and $2 a + 14 d = 44$ .
Answer
$4 d = 20$ , $d = 5$ , $a = - 13$ : the terms are −13, −8, −3.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the nth term of an A.P.Show answer
$T_{n} = a + (n - 1) d$
2.Find $d$ for 10, 7, 4, …Show answer
$- 3$
3.Find the 10th term of 2, 4, 6, …Show answer
$2 + 9 \times 2 = 20$
4.If $T_{n} = 5 n + 2$ , what is $d$ ?Show answer
5
5.When are $p$ , $q$ , $r$ in A.P.?Show answer
When $2 q = p + r$ .

nth Term of an A.P.: frequently asked questions

What is the formula for the nth term of an A.P.?

Tn = a + (n − 1)d, where a is the first term and d is the common difference. For example, in 4, 9, 14, … the 12th term is 4 + 11 × 5 = 59.

How do I check whether a number is a term of an A.P.?

Set a + (n − 1)d equal to the number and solve for n. If n is a positive whole number, the number is a term; otherwise it is not.

How do I find the A.P. when two terms are given?

Write each term as a + (position − 1)d to get two equations, subtract them to find d, and then find a.