Class 10Maths · AlgebraFull chapter

Arithmetic Progressions

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
What is an Arithmetic Progression?The nth Term of an APSum of the First n TermsSum Using the Last Term, and Linking Sₙ to aₙWord Problems on APs

What is an Arithmetic Progression?

Quick answer An AP is a list of numbers in which each term is obtained by adding a fixed number (the common difference d) to the previous term. The first term is a.

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always the same. This fixed difference is called the common difference and is denoted by d. The very first number in the list is called the first term, denoted by a.

If a list of terms is a1, a2, a3, a4, ..., then it is an AP only when a2 − a1 = a3 − a2 = a4 − a3 = d. So the common difference is found by subtracting any term from the term that comes right after it.

A general AP looks like: a, a + d, a + 2d, a + 3d, ...

Worked example: Consider the list 5, 8, 11, 14, 17, ...

  • First term a = 5
  • d = 8 − 5 = 3, and 11 − 8 = 3, and 14 − 11 = 3

Since the difference is the same (3) throughout, this is an AP with a = 5 and d = 3.

Note on d: The common difference can be positive (terms increase, e.g. 2, 5, 8, ...), negative (terms decrease, e.g. 10, 7, 4, ...) or zero (all terms equal, e.g. 4, 4, 4, ...). An AP with a finite number of terms is a finite AP and has a last term; an AP that continues forever is an infinite AP.

Common difference d = ak₊₁ − ak Difference between a term and the term just before it.
General form of an AP a, a + d, a + 2d, a + 3d, ... Built from first term a and common difference d.
Remember
  • Each term differs from the previous one by a fixed amount d.
  • a is the first term; d = a₂ − a₁ = any term minus its preceding term.
  • d can be positive, negative or zero.
  • General form: a, a + d, a + 2d, a + 3d, ...
  • To check if a list is an AP, verify that the difference between consecutive terms stays constant.

The nth Term of an AP

Quick answer The nth (general) term of an AP is aₙ = a + (n − 1)d, letting you find any term directly without listing all the earlier ones.

Writing out an AP shows a pattern in the terms: the 1st term is a, the 2nd is a + d, the 3rd is a + 2d, the 4th is a + 3d. Each time the number of d's added is one less than the term number. This gives the formula for the nth term (also called the general term):

an = a + (n − 1)d

Here an is the term in position n, a is the first term, d is the common difference and n is the number of the term. For a finite AP, the last term is often written as l, so l = a + (n − 1)d.

Worked example 1: Find the 11th term of the AP 5, 11, 17, 23, ...

  • a = 5, d = 11 − 5 = 6, n = 11
  • a11 = a + (11 − 1)d = 5 + 10 × 6 = 5 + 60 = 65

Worked example 2: Is 78 a term of the AP 3, 8, 13, 18, ...?

  • a = 3, d = 5. Set an = 78: 3 + (n − 1)5 = 78
  • (n − 1)5 = 75, so n − 1 = 15, giving n = 16 (a whole number)

Since n = 16 is a positive integer, 78 is the 16th term. If n had come out as a fraction, the number would not be a term of the AP.

nth term aₙ = a + (n − 1)d General term of an AP.
Last term l = a + (n − 1)d The last term of a finite AP with n terms.
nth term from the end = l − (n − 1)d Counting positions from the last term.
Remember
  • aₙ = a + (n − 1)d gives any term directly.
  • The last term of a finite AP is l = a + (n − 1)d.
  • To test if a number is a term, solve for n; it is a term only if n is a positive integer.
  • The number of d's added is always (n − 1), one less than the term position.
  • Two conditions (two given terms) let you find both a and d.

Sum of the First n Terms

Quick answer The sum of the first n terms of an AP is Sₙ = n/2[2a + (n − 1)d], obtained by pairing terms from the two ends.

Adding many terms of an AP one by one is slow, so we use a formula. If Sn denotes the sum of the first n terms, then:

Sn = n/2 [2a + (n − 1)d]

This comes from writing the sum forwards and backwards and adding the two copies together. That gives 2Sn, and every one of the n terms then pairs up to the same total, 2a + (n − 1)d. So 2Sn = n[2a + (n − 1)d], which after dividing by 2 gives the n/2 factor.

Worked example 1: Find the sum of the first 20 terms of the AP 1, 3, 5, 7, ...

  • a = 1, d = 2, n = 20
  • S20 = 20/2 [2 × 1 + (20 − 1) × 2] = 10 [2 + 38] = 10 × 40 = 400

Worked example 2: Find the sum of the first 15 terms of the AP 8, 5, 2, −1, ...

  • a = 8, d = −3, n = 15
  • S15 = 15/2 [2 × 8 + 14 × (−3)] = 15/2 [16 − 42] = 15/2 × (−26) = 15 × (−13) = −195

A special result: the sum of the first n positive integers 1 + 2 + 3 + ... + n = n(n + 1)/2, which is just this formula with a = 1 and d = 1.

Sum of n terms Sₙ = n/2 [2a + (n − 1)d] Using first term and common difference.
Sum of first n natural numbers 1 + 2 + ... + n = n(n + 1)/2 Special case a = 1, d = 1.
Remember
  • Sₙ = n/2[2a + (n − 1)d] adds the first n terms quickly.
  • Substitute a, d and n carefully, respecting the sign of d.
  • Works for increasing, decreasing or constant APs.
  • Sum of first n natural numbers: n(n + 1)/2.
  • Choose the correct n — the number of terms being added.

Sum Using the Last Term, and Linking Sₙ to aₙ

Quick answer When the last term l is known, Sₙ = n/2 (a + l); also each term can be recovered as aₙ = Sₙ − Sₙ₋₁.

When the last term l of a finite AP is known, the sum formula becomes simpler. Since l = a + (n − 1)d, we can replace 2a + (n − 1)d by a + l to get:

Sn = n/2 (a + l)

This says the sum equals the number of terms times the average of the first and last terms. Use this form whenever both a and l are given.

Worked example 1: Find the sum of the AP 7, 10, 13, ..., 97.

  • a = 7, d = 3, l = 97. First find n: 97 = 7 + (n − 1)3, so (n − 1)3 = 90, n − 1 = 30, n = 31
  • S31 = 31/2 (7 + 97) = 31/2 × 104 = 31 × 52 = 1612

There is also a useful link between the sum and the terms. Since Sn is the sum up to the nth term and Sn−1 is the sum up to the (n − 1)th term, their difference is exactly the nth term:

an = Sn − Sn−1

Worked example 2: The sum of the first n terms of an AP is Sn = 3n² + 2n. Find its 10th term.

  • a10 = S10 − S9
  • S10 = 3(100) + 2(10) = 320; S9 = 3(81) + 2(9) = 243 + 18 = 261
  • a10 = 320 − 261 = 59
Sum using last term Sₙ = n/2 (a + l) l is the last term of the finite AP.
nth term from sums aₙ = Sₙ − Sₙ−₁ Difference of consecutive partial sums.
Remember
  • Sₙ = n/2 (a + l) when the last term l is known.
  • Sum = (number of terms) × (average of first and last term).
  • Find n first if it is not given, using l = a + (n − 1)d.
  • aₙ = Sₙ − Sₙ−₁ recovers any term from the sum formula.
  • a₁ = S₁ is a quick check on the first term.

Word Problems on APs

Quick answer Real-life situations with a fixed step—savings, salaries, seating, production—are modelled as APs and solved with the nth-term and sum formulas.

Many everyday situations change by a fixed amount each step, so they form an AP. The skill is to read the problem, identify a (the starting value), d (the fixed change) and what is asked (a particular term an or a total Sn), and then apply the right formula.

Worked example 1 (finding a term): A company's production of televisions was 600 units in the 3rd year and 700 units in the 7th year. If production increases uniformly, find the production in the 10th year.

  • a3 = a + 2d = 600 and a7 = a + 6d = 700
  • Subtracting: 4d = 100, so d = 25, and a = 600 − 2(25) = 550
  • a10 = a + 9d = 550 + 9 × 25 = 550 + 225 = 775 units

Worked example 2 (finding a sum): A person saves ₹100 in the first month and increases the saving by ₹50 each month. How much is saved in 12 months?

  • a = 100, d = 50, n = 12
  • S12 = 12/2 [2 × 100 + (12 − 1) × 50] = 6 [200 + 550] = 6 × 750 = ₹4500

Tip: Decide whether the question asks for a single term (use an) or a running total (use Sn). Words like 'in the nth year' point to a term; words like 'total', 'altogether' or 'in n years' point to a sum.

Term needed aₙ = a + (n − 1)d For a value at a specific step.
Total needed Sₙ = n/2 [2a + (n − 1)d] For an accumulated total over n steps.
Remember
  • Translate the situation into a, d and n before calculating.
  • 'In the nth year/step' means find aₙ; 'total/altogether' means find Sₙ.
  • Two given conditions form two equations to get a and d.
  • Check that answers are sensible (whole numbers of items, etc.).
  • Common contexts: savings, salaries, production, seating rows, stacked logs.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

d = ak₊₁ − ak
Common difference
a, a + d, a + 2d, a + 3d, ...
General form of an AP
aₙ = a + (n − 1)d
nth term
l = a + (n − 1)d
Last term
= l − (n − 1)d
nth term from the end
Sₙ = n/2 [2a + (n − 1)d]
Sum of n terms
1 + 2 + ... + n = n(n + 1)/2
Sum of first n natural numbers
Sₙ = n/2 (a + l)
Sum using last term
aₙ = Sₙ − Sₙ−₁
nth term from sums
aₙ = a + (n − 1)d
Term needed
Sₙ = n/2 [2a + (n − 1)d]
Total needed

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Meaning of an AP easy

Which of the following lists of numbers is an Arithmetic Progression?

Q2 Common difference easy

The common difference of the AP 3, 1, −1, −3, ... is:

Q3 nth term formula easy

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

Q4 nth term medium

The 10th term of the AP 2, 7, 12, 17, ... is:

Q5 nth term medium

Which term of the AP 3, 7, 11, 15, ... is equal to 111?

Q6 Sum of n terms medium

The sum of the first n terms of an AP is given by:

Q7 Sum using last term medium

The sum of the AP 2, 4, 6, ..., 20 is:

Q8 Finding a and d hard

In an AP the 8th term is 31 and the 15th term is 16 more than the 11th term. The first term is:

Q9 Word/analysis problem medium

Which term of the AP 20, 17, 14, 11, ... is its first negative term?

Q10 Word problems hard

How many two-digit numbers are divisible by 3?

Q11 nth term medium

The 20th term of the AP with first term −3 and common difference 4 is:

Q12 Sum of n terms hard

If the sum of the first n terms of the AP 1, 4, 7, 10, ... is 590, then n equals:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Which term of the AP 21, 18, 15, ... is −81? Also, is any term of this AP equal to zero?nth term

Here a = 21 and d = 18 − 21 = −3.

The nth term is an = a + (n − 1)d = 21 + (n − 1)(−3) = 24 − 3n.

For an = −81: 24 − 3n = −81, so 3n = 105, giving n = 35.

Therefore −81 is the 35th term.

For a term equal to 0: 24 − 3n = 0, so 3n = 24, giving n = 8.

Since n = 8 is a positive integer, the 8th term is 0.

2 Which term of the AP 3, 8, 13, ..., 253 is the 20th term from the end?nth term from the end

Here a = 3, d = 5 and last term l = 253.

Method (from the end): The nth term from the end = l − (n − 1)d.

20th term from the end = 253 − (20 − 1)(5) = 253 − 19 × 5 = 253 − 95 = 158.

Check by position: Total terms: 253 = 3 + (n − 1)5 gives n = 51. The 20th from the end is the (51 − 20 + 1) = 32nd term = 3 + 31 × 5 = 158. This confirms the answer.

3 Find the sum of the first 22 terms of the AP 8, 3, −2, ...Sum of n terms

Here a = 8, d = 3 − 8 = −5 and n = 22.

Sn = n/2 [2a + (n − 1)d]

S22 = 22/2 [2 × 8 + (22 − 1)(−5)]

= 11 [16 + 21 × (−5)] = 11 [16 − 105] = 11 × (−89) = −979.

4 How many terms of the AP 24, 21, 18, ... must be taken so that their sum is 78?Sum of n terms

Here a = 24, d = −3 and Sn = 78.

Sn = n/2 [2a + (n − 1)d] = 78

n/2 [48 + (n − 1)(−3)] = 78

n/2 [48 − 3n + 3] = 78, so n(51 − 3n) = 156.

51n − 3n² = 156, i.e. 3n² − 51n + 156 = 0, or n² − 17n + 52 = 0.

Factorising: (n − 4)(n − 13) = 0, so n = 4 or n = 13.

Both values are valid positive integers, so either 4 terms or 13 terms give the sum 78. (This happens because the later terms become negative and cancel out, e.g. the 5th to 13th terms sum to zero.)

5 Determine the AP whose 3rd term is 16 and whose 7th term exceeds the 5th term by 12.Finding a and d

Let the first term be a and common difference d.

Condition 1: a3 = a + 2d = 16.

Condition 2: a7 − a5 = 12, i.e. (a + 6d) − (a + 4d) = 12, so 2d = 12, giving d = 6.

Substituting in Condition 1: a + 2(6) = 16, so a = 16 − 12 = 4.

Therefore the AP is 4, 10, 16, 22, 28, ...

6 If the sum of the first 7 terms of an AP is 49 and that of the first 17 terms is 289, find the sum of its first n terms.Sum of n terms

Using Sn = n/2 [2a + (n − 1)d]:

S7 = 49: 7/2 [2a + 6d] = 49, so 7(a + 3d) = 49, giving a + 3d = 7. ...(i)

S17 = 289: 17/2 [2a + 16d] = 289, so 17(a + 8d) = 289, giving a + 8d = 17. ...(ii)

Subtracting (i) from (ii): 5d = 10, so d = 2. Then a = 7 − 3(2) = 1.

Sum of first n terms: Sn = n/2 [2(1) + (n − 1)(2)] = n/2 [2 + 2n − 2] = n/2 × 2n = n².

Previous-year board questions 4

Q1 The nth term of an AP is given by aₙ = 7 − 4n. Find its common difference. CBSE 2023 1 mark

Substitute values of n to get the terms.

a1 = 7 − 4(1) = 3 and a2 = 7 − 4(2) = −1.

Common difference d = a2 − a1 = −1 − 3 = −4.

(In general, when an is a linear expression in n, the coefficient of n is the common difference, here −4.)

Q2 Which term of the AP 21, 42, 63, 84, ... is 210? CBSE 2020 2 marks

Here a = 21 and d = 42 − 21 = 21.

Let the nth term be 210: an = a + (n − 1)d = 210.

21 + (n − 1)(21) = 210

(n − 1)(21) = 189, so n − 1 = 9, giving n = 10.

Therefore 210 is the 10th term of the AP.

Q3 Find the sum of the first 40 positive integers divisible by 6. CBSE 2022 3 marks

The positive integers divisible by 6 are 6, 12, 18, 24, ... — an AP with a = 6 and d = 6.

The 40th term (last term): l = a + (n − 1)d = 6 + (40 − 1)(6) = 6 + 234 = 240.

Sum: S40 = n/2 (a + l) = 40/2 (6 + 240) = 20 × 246 = 4920.

Q4 The sum of the first 14 terms of an AP is 1050 and its first term is 10. Find the 20th term of the AP. CBSE 2019 5 marks

Given a = 10 and S14 = 1050.

Sn = n/2 [2a + (n − 1)d]

1050 = 14/2 [2(10) + (14 − 1)d] = 7 [20 + 13d]

So 20 + 13d = 1050 / 7 = 150, giving 13d = 130, so d = 10.

Now the 20th term: a20 = a + 19d = 10 + 19 × 10 = 10 + 190 = 200.

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