Arithmetic Progression — Class 10 Notes

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Maths · Ch 5Class 10 · CBSE 2026–27High weightage

Arithmetic Progressions

What the board actually asks: finding the nth term (or which term / how many terms), the sum of the first n terms, and modelling a real situation as an AP -- these three carry almost every mark here.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

Add the same step, again and again.

An arithmetic progression is a list of numbers where each term is the previous one plus a fixed step $d$ (the common difference). Pin down just two numbers -- the first term $a$ and the common difference $d$ -- and the whole list is fixed: $a, a + d, a + 2 d, \dots$ . The chapter answers three questions: what is the nth term? ( $a_{n} = a + (n - 1) d$ ), which term / how many terms? (solve that formula for $n$ ), and what is the sum of the first n terms? ( $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ ).

Formulae

General form

a, a + d, a + 2 d, a + 3 d, \dots

Tip: a, a+d, a+2d, …

Common difference

d = a_{k + 1} - a_{k}

Tip: d = a₂ − a₁; every consecutive difference must be equal. For a formula, show aₙ₊₁ − aₙ is constant.

nth term

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

Tip: aₙ = a + (n−1)d; find n by solving for it.

Sum of n terms

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

Tip: Use this when d is known.

Sum with last term l

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

Tip: Use this when the last term is given.

nth Term from the Sum

a_{n} = S_{n} - S_{n - 1}

for

n \geq 2

, and

a_{1} = S_{1}

mth Term from the End

l

is the last term, the

m

th term from the end is

l - (m - 1) d

How to Tell If a List is an AP

A list is an AP if the difference between every pair of consecutive terms is the same.

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Arithmetic Progression (AP)

NCERT verbatim — memorise this exact wording· Ch 5,

“So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.”

What it means: A running list where you reach the next number by adding the SAME fixed amount every time. To test a list, check that consecutive differences are all equal.

Common difference

NCERT verbatim — memorise this exact wording· Ch 5,

“This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.”

What it means:

d = a_{k + 1} - a_{k}

-- subtract any term from the one AFTER it (not before). The sign of

d

tells you whether the AP rises (

d > 0

), falls (

d < 0

) or stays flat (

d = 0

General form of an AP

NCERT verbatim — memorise this exact wording· Ch 5,

“a, a + d, a + 2d, a + 3d, . . . represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.”

What it means: Every AP is captured by two numbers: the first term

a

and the common difference

d

. Know those and you can write the whole list.

nth term of an AP

NCERT verbatim — memorise this exact wording· Ch 5,

“So, the nth term an of the AP with first term a and common difference d is given by an = a + (n – 1) d.”

What it means:

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

-- take

(n - 1)

steps of size

d

from the first term. This single formula answers 'find the 20th term', 'which term is 78?', and 'how many terms?' (rearrange for

n

Sum of the first n terms

NCERT verbatim — memorise this exact wording· Ch 5,

“So, the sum of the first n terms of an AP is given by S = n/2 [2a + (n – 1) d]”

What it means:

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

. If the last term

l

is already known, the shortcut

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

is faster. Note

S_{n}

needs the number of terms

n

, not a term's value.

Key terms — NCERT verbatim (state word-for-word)

Term: Each of the numbers in the list is called a term. (Ch 5, )
Finite AP: Note that in examples (a) to (e) above, there are only a finite number of terms. Such an AP is called a finite AP. (Ch 5, )
Infinite AP: The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term. (Ch 5, )
General term and last term (l): an is also called the general term of the AP. If there are m terms in the AP, then am represents the last term which is sometimes also denoted by l. (Ch 5, )
Arithmetic mean: If a, b, c are in AP, then b = (a + c)/2 and b is called the arithmetic mean of a and c. (Ch 5, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Recognise an AP

A list is an AP only if the gap between consecutive terms is constant:

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

. Compute one or two differences before assuming -- squares (

1, 4, 9, \dots

) and doubling lists (

2, 4, 8, \dots

) are the classic traps that look regular but are NOT APs.

FigureAn AP on the number line -- every hop adds the same common difference d

Figure asset pending extraction — the sourced original is Figure.

Terms a, a+d, a+2d, . . . sit at equal gaps of d. A constant hop is exactly what makes a list an AP; change the hop and it stops being one.

MCQ / 'which of these forms an AP'

The nth term

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

. Read off the first term

a

and common difference

d

, then substitute the position

n

. This is the most-used tool in the chapter.

1-2 mark 'find the term'

Sum of the first n terms

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

, or

\frac{n}{2} (a + l)

when the last term

l

is known. The formula links four quantities

S_{n}, a, d, n

-- given any three you can find the fourth.

3 mark 'find the sum'

Solve for n -- which term / how many terms

Set

a_{n}

(or

S_{n}

) equal to the target value and solve for

n

n

MUST come out a positive integer -- a fractional or negative

n

means the value is simply not a term of the AP. A sum condition gives a quadratic in

n

, which can yield two valid answers.

2-3 mark

Model a real situation as an AP

A fixed-increment salary, uniform monthly savings, rows of plants decreasing by a constant, a penalty rising by a fixed step -- extract the first term

a

and the constant change

d

, then apply the term or the sum formula. Reject any answer the situation forbids (you cannot have a fractional number of rows).

3-5 mark applied / case-based

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 5, Example 3,

“Find the 10th term of the AP : 2, 7, 12, . . .”

Read off a, d, n and write the formula.

a = 2

d = 7 - 2 = 5

n = 10

; the nth term is

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

.Concept: identify a and d

1 mark

Substitute and evaluate.

a_{10} = 2 + (10 - 1) \times 5 = 2 + 45 = 47

. So the 10th term is

47

.Concept: substitute into the nth-term formula

1 mark

Mark-scheme logic (authored): 2-mark shape (Section B): 1 for the correct data + formula

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

, 1 for the signed substitution and the answer.

Ch 5, Example 4,

“Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.”

Ch 5, Example 11,

“Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .”

Ch 5, Example 13,

“How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?”

Ch 5, Example 10,

“In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?”

Ch 5, Example 16,

“A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find : (i) the production in the 1st year (ii) the production in the 10th year (iii) the total production in first 7 years”

Problem-solving strategies AUTHORED

Given a list -- is it an AP?Check the differences:

a_{2} - a_{1}

a_{3} - a_{2}

. Equal every time means AP with that value as

d

; one mismatch means it is not.

Find a particular term?Read off

a

and

d

, then

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

Which term / how many terms?Set

a_{n}

(or

S_{n}

) equal to the target and solve for

n

n

must be a positive integer.

Sum asked?

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

; if the last term

l

is known,

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

is quicker.

Word problem?Pull the first term

a

and the fixed change

d

from the situation, apply the term or sum formula, then reject any impossible answer.

Common mistakes that cost marks

Where students lose easy marks

Subtracting the wrong way for

d

-- always (later term

-

earlier term). For

6, 3, 0, - 3, \dots

the common difference is

- 3

, not

+ 3

Confusing a term's value with its position

n

-- 'which term is 78?' asks for

n

, not for a value.

Putting a term value where

n

belongs inside

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

-- the sum formula uses the number of terms.

Forgetting

n

must be a positive integer -- a fractional or negative

n

means the number is NOT a term of the AP.

Mixing the two sum forms --

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

needs

a

and

d

;

\frac{n}{2} (a + l)

needs the last term

l

Formulae

General form

a, a + d, a + 2 d, a + 3 d, \dots

Tip: a, a+d, a+2d, …

Common difference

d = a_{k + 1} - a_{k}

Tip: d = a₂ − a₁; every consecutive difference must be equal. For a formula, show aₙ₊₁ − aₙ is constant.

nth term

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

Tip: aₙ = a + (n−1)d; find n by solving for it.

Sum of n terms

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

Tip: Use this when d is known.

Sum with last term l

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

Tip: Use this when the last term is given.

nth Term from the Sum

a_{n} = S_{n} - S_{n - 1}

for

n \geq 2

, and

a_{1} = S_{1}

mth Term from the End

l

is the last term, the

m

th term from the end is

l - (m - 1) d

How to Tell If a List is an AP

A list is an AP if the difference between every pair of consecutive terms is the same.

Shortcuts & Tricks

aₙ = Sₙ − Sₙ₋₁ (and a₁ = S₁).

Safe in written answers

Last term given → n/2(a+l). Otherwise → n/2[2a+(n−1)d].

MCQ / check only: don't write this as working

mth term from the end = l − (m−1)d.

Safe in written answers

d = (aₘ − aₙ)/(m − n) when two terms are known.

MCQ / check only: don't write this as working

n must be a positive whole number; if it comes out fractional or negative, re-check or reject.

Safe in written answers

Memory Tricks

Pehla + (n−1) kadam.

aₙ = a + (n−1)d. Pehla = a; (n−1) = number of steps; kadam = d.

Aadha-n × (pehla + aakhri).

Sₙ = n/2(a + l). Aadha-n = n/2; pehla = a; aakhri = l.

Aaj tak ka total − kal tak ka total = aaj ka term.

aₙ = Sₙ − Sₙ₋₁. aaj tak = Sₙ; kal tak = Sₙ₋₁.

Mindmap

- - Constant common difference d
  - d = a(k+1) - a(k), any consecutive pair
  - d can be +, - or 0
  - General form a, a+d, a+2d, . . .
- - a(n) = a + (n-1)d
  - general term; last term = l
  - find a specific term
- - S(n) = n/2 [2a + (n-1)d]
  - S(n) = n/2 (a + l) when l known
  - links S, a, d, n -- any 3 give the 4th
- - which term / how many terms
  - set a(n) or S(n) = target
  - n must be a positive integer
  - sum → quadratic in n → two answers possible
- - salary / savings / rows / penalty
  - pull a and d from the situation
  - apply term or sum formula
  - reject the impossible answer
- - wrong-way subtraction for d
  - term value vs position n
  - n must be a whole number
  - pick the right sum form

Sum of the first n terms -- where the formula comes from

NCERT motivates the sum formula with Gauss's schoolboy trick (p.63): to add 1 to 100 he wrote the list forwards, then backwards, and added the two lines so every column gave the same total 101. The same reversal turns the sum of ANY AP into n identical pairs -- that is the whole derivation.

Write the sum of the first n terms.

S = a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d]

Write the same sum again, terms reversed.

S = [a + (n - 1) d] + [a + (n - 2) d] + \dots + (a + d) + a

Add the two lines term by term -- each of the n column-pairs gives the same total.

2 S = [2 a + (n - 1) d] + \dots + [2 a + (n - 1) d] (n times)

First+last, second+second-last, and so on each equal 2a + (n-1)d.

There are n equal pairs.

2 S = n [2 a + (n - 1) d]

Divide by 2 -- the sum formula.

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

Regroup a + (n-1)d as the last term to get the a-and-l form.

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

Useful when the first term a and last term l are known but d is not (NCERT form (4), p.64).

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

5 marksSection D

Jaspal Singh repays a total loan of ₹1,18,000 by paying every month, starting with the first instalment of ₹1000, increasing by ₹100 every month. What amount will he pay in the 30th instalment? After paying the 30th instalment, how much loan does he still have to pay?

Step-marked solution

[1 mark] The monthly instalments form an AP with first term a = 1000 and common difference d = 100.
[1 mark] 30th instalment a₃₀ = a + 29d = 1000 + 29 × 100 = 1000 + 2900 = ₹3900.
[1 mark] Total repaid in 30 instalments S₃₀ = (30/2)[2a + 29d] = 15[2 × 1000 + 29 × 100].
[1 mark] S₃₀ = 15 × (2000 + 2900) = 15 × 4900 = ₹73,500.
[1 mark] Loan still to be paid = 1,18,000 − 73,500 = ₹44,500.

5 marksSection D

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

Step-marked solution

[1 mark] S₅ = (5/2)(2a + 4d) = 5(a + 2d) and S₇ = (7/2)(2a + 6d) = 7(a + 3d).
[1 mark] S₅ + S₇ = (5a + 10d) + (7a + 21d) = 12a + 31d = 167.
[1 mark] S₁₀ = (10/2)(2a + 9d) = 10a + 45d = 235, i.e. 2a + 9d = 47.
[1 mark] Solve 12a + 31d = 167 with 2a + 9d = 47 (×6 ⇒ 12a + 54d = 282): subtract to get 23d = 115 ⇒ d = 5, then a = 1.
[1 mark] S₂₀ = (20/2)(2a + 19d) = 10(2 + 95) = 970.

5 marksSection D

An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

Step-marked solution

[1 mark] With 37 terms the middle term is the 19th; the three middle terms are a₁₈, a₁₉, a₂₀ with sum 3a₁₉ = 225, so a₁₉ = 75, i.e. a + 18d = 75.
[1 mark] The last three terms a₃₅, a₃₆, a₃₇ have sum 3a₃₆ = 429, so a₃₆ = 143, i.e. a + 35d = 143.
[1 mark] Subtract the two equations: 17d = 68.
[1 mark] d = 4, and a = 75 − 18 × 4 = 3.
[1 mark] The AP is 3, 7, 11, 15, ...

Practise Arithmetic Progressions competency questions

Read NCERT Chapter 5 on LazyTopper (jemh105)

Official NCERT chapter PDF (jemh105) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 5 (jemh105.pdf, printed pages 49-72). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Arithmetic Progression (AP)	verbatim	NCERT 2026-27 Ch5
Common difference	verbatim	NCERT 2026-27 Ch5
General form of an AP	verbatim	NCERT 2026-27 Ch5
nth term of an AP	verbatim	NCERT 2026-27 Ch5
Sum of the first n terms	verbatim	NCERT 2026-27 Ch5
Term	verbatim	NCERT 2026-27 Ch5
Finite AP	verbatim	NCERT 2026-27 Ch5
Infinite AP	verbatim	NCERT 2026-27 Ch5
General term and last term (l)	verbatim	NCERT 2026-27 Ch5
Arithmetic mean	verbatim	NCERT 2026-27 Ch5 (A Note to the Reader)
Find the nth term . NCERT Example 3	example	NCERT 2026-27 Ch5 Example 3
Which term is -81? Is any term 0? . NCERT Example 4	example	NCERT 2026-27 Ch5 Example 4 -59
Sum of first n terms . NCERT Example 11	example	NCERT 2026-27 Ch5 Example 11
How many terms give a required sum . NCERT Example 13	example	NCERT 2026-27 Ch5 Example 13
Word problem -- how many rows . NCERT Example 10	example	NCERT 2026-27 Ch5 Example 10
Multi-part word problem -- production . NCERT Example 16	example	NCERT 2026-27 Ch5 Example 16 -68

Arithmetic Progressions · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 5 (jemh105.pdf, printed pages 49-72)

Practise writing answers the way examiners mark them → Guide