Class 10 Maths Notes

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Arithmetic Progressions Class 10 Maths Notes

An arithmetic progression is any list in which every term is the previous term plus the same fixed number, and that single idea is the whole chapter. Once you can name the common difference d you can reach any term with aₙ = a + (n − 1)d, and once you have the first and the last term you can add the whole list in one line with Sₙ = n/2 (a + l).

Class:10Subject:MathematicsUnit:IICovers:CBSE 2024-25

7 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

What are the formula for the nth term and the formula for the sum of the first n terms of an arithmetic progression?

The nth term is aₙ = a + (n − 1)d, where a is the first term and d the common difference. The sum of the first n terms is Sₙ = n/2 [2a + (n − 1)d], which is the same thing written with the first term and the common difference. Equivalently, if l is the last term aₙ then Sₙ = n/2 (a + l), and a + l = 2a + (n − 1)d, so the two forms always agree.

Why a Sequence Is Studied at All

The board motivates this section with the observation that quantities in daily life very often change by a fixed amount. Rent goes up by the same sum every year, a salary rises by the same amount each year, a queue grows by one more person in every successive row, a staircase gets one step taller at each level. When the step is constant the numbers form an arithmetic progression, and once they do, two formulas let you reach any term or the whole sum without writing the list out.

A salary that rises by ₹500 every year, a deposit that earns a fixed interest every year, a bus route whose stages get longer by a fixed distance.
A queue of children in rows, where each row has a fixed number more children than the row in front.
A staircase where each step is a fixed number of bricks higher than the step below it.
The board's advice is short and worth following: notice the pattern first, name a and d, and then choose between the two sum forms depending on whether you know the last term.

Where this sits in the course

Arithmetic Progressions is the last section of Unit II Algebra, which carries 20 of the 80 written marks in the Class 10 paper. The 2024-25 course asks for the motivation, the derivation of the nth term and of the sum of the first n terms, and their application in solving daily-life problems. That is the whole scope.

What an Arithmetic Progression Is

Definition

Arithmetic progression

A list of numbers a₁, a₂, a₃, ... is an arithmetic progression, or an AP, if the difference between any term and the term immediately before it is constant. That constant is the common difference, written d, so a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = d. The common difference may be positive, zero or negative.

The definition of an AP: every gap is the same gap.

a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = d

5, 11, 17, 23, ... is an AP with a = 5 and d = 6, since 11 − 5 = 6, 17 − 11 = 6 and 23 − 17 = 6.
9, 6, 3, 0, −3, ... is an AP with a = 9 and d = −3. A negative common difference is perfectly normal.
4, 4, 4, 4, ... is an AP with d = 0. This one is quietly in the syllabus and is often left out of practice sets.
2, 6, 10, 14, 18, ... is an AP with d = 4.
1, 4, 7, 13, 16, ... is NOT an AP, because the gaps are 3, 3, 6 and 3.

The thirty-second test for an AP

Write the consecutive differences underneath the list and check that they are all equal. If even one gap is different, the list is not an AP, whatever else is true about it. In the examination you are usually asked only this, and one line of differences settles it.

The nth Term

The first term is a₁ = a. Each step adds d, so to reach the nth term you take n − 1 steps, not n. That single off-by-one is the most common error in the chapter, so build the derivation from a₁ + (n − 1)d and never from memory.

The nth term of an AP with first term a and common difference d.

a_{n} = a + (n - 1) d

In 5, 11, 17, 23, ... with a = 5 and d = 6, the 15th term is a₁₅ = 5 + 14 × 6 = 5 + 84 = 89.
Is 47 a term of 5, 11, 17, 23, ...? Put aₙ = 47, so 5 + (n − 1) × 6 = 47, which gives 6(n − 1) = 42 and n − 1 = 7, so n = 8. Yes, 47 is the 8th term.
In 9, 6, 3, 0, −3, ... the 20th term is a₂₀ = 9 + 19 × (−3) = 9 − 57 = −48.
If the 7th term is 34 and the 13th term is 64, subtract the two: a₁₃ − a₇ = (a + 12d) − (a + 6d) = 6d, so 6d = 64 − 34 = 30 and d = 5. Then a = a₇ − 6d = 34 − 30 = 4, and a₁₉ = 4 + 18 × 5 = 94.

Count the steps, not the terms

From the 1st term to the 7th term there are 6 steps, not 7, so a + 6d and not a + 7d. If you write a₇ = a + 7d every time, the answer to the 15th-term question comes out as 95 instead of 89 and the error is invisible until the checking stage.

Working backwards: the common difference from two known terms, the nth term from one known term, and the number of terms.

d = \frac{a _{p} - a _{q}}{p - q}, a_{n} = a_{p} + (n - p) d, n = \frac{a _{n} - a _{1}}{d} + 1

The Sum of the First n Terms

The board asks you to derive Sₙ, and the derivation is the pairing argument: write the sum forwards, write it backwards, add the two lines, and every column gives the same total a + l. That is where the factor n/2 comes from, and it is why the sum only needs the first and the last term.

The three equal ways to write Sₙ. Define l as the last term, so l = aₙ.

S_{n} = \frac{n}{2} (a + l) = \frac{n}{2} (a + a_{n}) = \frac{n}{2} [2 a + (n - 1) d]

The first form, Sₙ = n/2 (a + l), is the one to use whenever you have the last term, or whenever you can work out the last term quickly.
The second form, Sₙ = n/2 [2a + (n − 1)d], is the one to use when you know a and d but not the last term. Since a + l = a + a + (n − 1)d = 2a + (n − 1)d, the two forms are the same number every time.
The average of the first n terms is (a + l)/2, because the sum is n/2 (a + l) and there are n terms. This is the mean of an AP and it is the mean of the first and the last term.
If you are ever unsure which form to use, use Sₙ = n/2 [2a + (n − 1)d]. It needs nothing but a, d and n, and every one of them is given or derivable in most questions.

Write down what l means before you use it

In this chapter l means the last term of the first n terms, that is l = aₙ. Students who pick up a formula from another book and meet l used for something else end up with the right structure and the wrong number. Define it yourself on the first line of the answer: l = aₙ.

A Worked Sum Problem with a Whole-Number Answer

Find the sum of the first 40 terms of the AP 3, 7, 11, .... Here a = 3, d = 4 and n = 40. Work out the last term first, then use the a + l form, and confirm with the 2a + (n − 1)d form so that you know the two agree.

a = 3, d = 4, n = 40. First the last term: a₄₀ = 3 + (40 − 1) × 4 = 3 + 39 × 4 = 3 + 156 = 159. So l = 159.
Now the first form: S₄₀ = 40/2 × (3 + 159) = 20 × 162 = 3240.
Now the second form, as a check: S₄₀ = 40/2 × [2 × 3 + (40 − 1) × 4] = 20 × [6 + 156] = 20 × 162 = 3240. The same answer.
So the sum of the first 40 terms is 3240.

Both forms of Sₙ on one line, agreeing at 3240.

S_{40} = \frac{40}{2} (3 + 159) = \frac{40}{2} [6 + 39 \times 4] = 20 \times 162 = 3240

Which form saves time

If you are going to need the last term for anything else in the question, find it once and use Sₙ = n/2 (a + l). If the last term is not wanted, go straight to Sₙ = n/2 [2a + (n − 1)d] and save the line of working.

The classic how-many-terms question

How many terms of the AP 24, 21, 18, ... must be taken so that the sum is 78? Here a = 24 and d = −3, so 78 = n/2 [2 × 24 + (n − 1)(−3)] = n/2 [48 − 3n + 3] = n(51 − 3n)/2. Then 156 = 51n − 3n², so 3n² − 51n + 156 = 0 and n² − 17n + 52 = 0, that is (n − 13)(n − 4) = 0. Both n = 13 and n = 4 work: the first 4 terms 24, 21, 18, 15 add to 78, and S₁₃ = 13/2 × [48 − 36] = 78 as well.

Daily-Life Applications

A daily-life AP problem gives you a situation and asks for a term, a sum, a cost or a count. The method never changes: name the first term a and the common difference d from the words of the question, identify n, and apply the formula the question needs. The only real skill is reading the situation correctly enough to spot which of a and d is being described.

Savings: a man saves ₹100 in the first week, ₹150 in the second and ₹200 in the third, and so on. Here a = 100 and d = 50. In the 20th week he saves a₂₀ = 100 + 19 × 50 = 100 + 950 = ₹1050.
Savings, the sum: how much does he save in 20 weeks? l = 1050, so S₂₀ = 20/2 × (100 + 1050) = 10 × 1150 = ₹11500.
Queue: in an assembly the first row has 12 children and each row behind it has 3 more. The 15th row has a₁₅ = 12 + 14 × 3 = 12 + 42 = 54 children, and the first 15 rows together hold S₁₅ = 15/2 × (12 + 54) = 15 × 33 = 495 children.
Staircase: a staircase has 8 steps, the first step being 12 bricks high and each step above it 2 bricks higher. The heights 12, 14, 16, ... form an AP with a = 12 and d = 2, so a₈ = 12 + 7 × 2 = 26 and the total is S₈ = 8/2 × (12 + 26) = 4 × 38 = 152 bricks.
Logs: a stack of logs has 20 logs in the bottom row, 19 in the next and so on up to 1 at the top. That is an AP with a = 20, d = −1 and n = 20, so S₂₀ = 20/2 × (20 + 1) = 10 × 21 = 210 logs.

Always state a, d and n before the formula

Three short lines, Then the formula. In every one of these examples a, d and n come straight from the wording, so writing them out first removes the most common source of a wrong answer, which is using a plausible-looking but unstated value for one of them.

How the Questions Are Asked

The questions in this section divide cleanly into four types: test whether a list is an AP, find a missing term, find the sum of the first n terms, and solve a daily-life problem. The first is one mark, the second two, the third three and the last five, so the long question is always a sum or a nth-term question dressed up in a story.

Check whether the list 1, 4, 7, 13, 16 is an arithmetic progression. (1 to 2 marks.)
Find the 15th term of the AP 5, 11, 17, 23, .... (2 marks, Section B.)
Find the sum of the first 40 terms of the AP 3, 7, 11, .... (3 marks, Section C.)
How many terms of the AP 24, 21, 18, ... must be taken so that the sum is 78? (5 marks, Section D.)
If the 7th term of an AP is 34 and the 13th term is 64, find its 19th term. (5 marks.)
A man saves ₹100 in the first week and ₹50 more every week after that. How much does he save in 20 weeks? (5 marks, Section D.)
Which term of the AP 5, 11, 17, 23, ... is 47? (2 marks.)

Paper reminders

The Class 10 paper is 80 marks in 3 hours with 38 questions. Q1-18 are one-mark MCQs and Q19-20 are one-mark assertion-reason, Q21-25 carry 2 marks, Q26-31 carry 3 marks, Q32-35 carry 5 marks and Q36-38 are 4-mark case studies, with internal choice in two questions each of Sections B, C and D. Unit II Algebra carries 20 of the 80 written marks, and the other 20 marks are internal assessment, 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical. Take π = 22/7 unless the question says otherwise, and calculators are not allowed.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

nth term of an AP

a_{n} = a + (n - 1) d

From the first term to the nth there are n − 1 steps, not n.

Sum using the first and last term

S_{n} = \frac{n}{2} (a + l), l = a_{n}

l means the last term of the first n terms. The mean of the terms is (a + l)/2.

Sum using a and d only

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

The same sum as n/2 (a + l), because a + l = 2a + (n − 1)d.

Common difference from two terms

d = \frac{a _{p} - a _{q}}{p - q}

Divides the difference of the values by the difference of the positions.

nth term from a known term

a_{n} = a_{p} + (n - p) d

Useful when the first term of the AP is not given.

Number of terms

n = \frac{a _{n} - a _{1}}{d} + 1

Turns a statement about which term a value is into its position.

The three sum forms together

\frac{n}{2} (a + l) = \frac{n}{2} (a + a_{n}) = \frac{n}{2} [2 a + (n - 1) d]

All three are identical. Use whichever needs the fewest extra steps.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

A list is an AP if and only if all the consecutive differences are equal. 1, 4, 7, 13, 16 is not an AP, because the gaps are 3, 3, 6, 3.
aₙ = a + (n − 1)d. From the 1st term to the 7th term there are 6 steps, so it is a + 6d and not a + 7d.
Sₙ = n/2 (a + l) with l = aₙ, and Sₙ = n/2 [2a + (n − 1)d]. These are the same number, so never print both as if they were different answers.
The mean of an AP is (a + l)/2, the mean of its first and last term — useful when a question asks for an average.
d = (aₚ − a_q)/(p − q). If the 7th term is 34 and the 13th is 64, then 6d = 30, so d = 5 and a = 4, giving a₁₉ = 94.
In a daily-life problem, write a, d and n on three separate lines before the formula. Almost every wrong answer in this section comes from a plausible but unstated value for one of them.
A negative common difference is normal. 24, 21, 18, ... has d = −3, and 20, 19, 18, ... is a stack of logs with a = 20 and d = −1.
Do not build answers on exercises built round a fixed sum of n terms. The course asks for the derivation of the formula and its use in daily-life problems, which is what these sections give.
Unit II Algebra carries 20 of the 80 written marks; the remaining 20 marks are internal assessment, 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical.

FAQ

Frequently asked questions

What is an arithmetic progression and what is its common difference?

A list of numbers is an arithmetic progression if the difference between every term and the term before it is the same fixed number. That number is the common difference d, so a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = d. For 5, 11, 17, 23 the gaps are all 6, so d = 6. The common difference can be positive, negative or zero, and 4, 4, 4 is an AP with d = 0.

What are the formulas for the nth term and the sum of the first n terms?

The nth term is aₙ = a + (n − 1)d. The sum of the first n terms is Sₙ = n/2 [2a + (n − 1)d], and if l is the last term aₙ then Sₙ = n/2 (a + l). The two sums are identical because a + l = 2a + (n − 1)d. For the AP 3, 7, 11, ... and n = 40, a₄₀ = 159 and S₄₀ = 20 × (3 + 159) = 3240.

How do I check whether a given list is an AP?

Subtract each term from the one after it and check that all the gaps are equal. For 2, 6, 10, 14, 18 the gaps are 4, 4, 4, 4, so it is an AP with d = 4. For 1, 4, 7, 13, 16 the gaps are 3, 3, 6, 3, so it is not an AP. A single unequal gap is enough to disqualify the list.

How is an AP used in a daily-life problem?

Read the situation for the fixed step, name it d, and take the first term a from the wording. A man saving ₹100 in the first week and ₹50 more each week after that gives a = 100 and d = 50, so his 20th week's saving is 100 + 19 × 50 = ₹1050 and the total of 20 weeks is 20/2 × (100 + 1050) = ₹11500. Rows of a queue, staircases and stacks of logs are all the same idea.

How do I find which term of an AP a given number is?

Set aₙ equal to that number and solve for n. For 47 in the AP 5, 11, 17, 23, ... we get 5 + (n − 1) × 6 = 47, so 6(n − 1) = 42 and n − 1 = 7, giving n = 8. So 47 is the 8th term. Remember that the position is one more than the number of steps.

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