Class 10 Maths Chapter 5: Arithmetic Progressions

Arithmetic Progressions is Chapter 5 of NCERT Class 10 Maths. It deals with number patterns in which each term goes up or down by the same fixed amount, and gives you formulas for any term and for the sum of many terms. Watch the four video lessons below, then use these notes, formulas and solved examples to practise.

What you will learn

• What makes a list of numbers an arithmetic progression (AP) and how to find its common difference
• The formula for the nth term and how to check whether a number belongs to an AP
• The formula for the sum of the first n terms
• Finding terms from the end and working backwards from given terms
• Applying AP formulas to real-life situations such as savings, instalments and seating rows

Video Lessons

Arithmetic Progressions - Part 1

Arithmetic Progressions - Part 2

Arithmetic Progressions - Part 3

Arithmetic Progressions - Part 4

Key Concepts and Formulas

What Is an AP

An arithmetic progression is a list of numbers where every term after the first is obtained by adding the same fixed number to the previous term. This fixed number can be positive, negative or zero.

Terms: a₁, a₂, a₃, ... (first term a = a₁)
d = a₂ − a₁ = a₃ − a₂ = ...

• If d > 0 the AP is increasing, if d < 0 it is decreasing, and if d = 0 all terms are equal
• An AP with a last term is finite; one that goes on forever is infinite

The nth Term

You do not need to write out every term to find, say, the 50th one. The general term follows directly from the pattern a, a + d, a + 2d, and so on.

aₙ = a + (n − 1)d
nth term from the end = l − (n − 1)d
Three terms in AP: a − d, a, a + d
a, b, c in AP ⇔ 2b = a + c

• To test whether a number p is a term, solve p = a + (n − 1)d; it is a term only if n is a positive integer

Sum of First n Terms

The sum formula comes from writing the series forwards and backwards and adding the two, which pairs up terms with the same total.

Sₙ = (n/2)[2a + (n − 1)d]
Sₙ = (n/2)(a + l), where l = last term
aₙ = Sₙ − Sₙ₋₁
1 + 2 + ... + n = n(n + 1)/2

Choosing the Right Formula

In the CBSE board exam, most AP questions are decided by picking the right formula quickly.

• Asked for a particular term → use the nth term formula
• Asked for a total, and d is known → use the sum formula with 2a + (n − 1)d
• Asked for a total, and the last term is known → use the sum formula with a + l
• Given two terms → write two equations in a and d and solve them together

Solved Examples

Example 1: Find the 15th term of the AP 4, 9, 14, 19, ...

a = 4, d = 9 − 4 = 5, n = 15
a₁₅ = 4 + (15 − 1) × 5
a₁₅ = 4 + 70
∴ a₁₅ = 74

Example 2: Which term of the AP 7, 11, 15, ... is 87?

a = 7, d = 4, aₙ = 87
87 = 7 + (n − 1) × 4
(n − 1) × 4 = 80
n − 1 = 20
∴ n = 21, so 87 is the 21st term

Example 3: Is 150 a term of the AP 5, 12, 19, ...?

a = 5, d = 7
150 = 5 + (n − 1) × 7
(n − 1) × 7 = 145
n − 1 = 145/7 (not a whole number)
∴ 150 is not a term of this AP

Example 4: Find the sum of the first 20 terms of 3, 8, 13, ...

a = 3, d = 5, n = 20
S₂₀ = (20/2)[2 × 3 + (20 − 1) × 5]
S₂₀ = 10 × (6 + 95)
∴ S₂₀ = 1010

Example 5: Find the sum of all two-digit multiples of 6.

AP: 12, 18, ..., 96 (a = 12, d = 6, l = 96)
96 = 12 + (n − 1) × 6
n − 1 = 14, so n = 15
S = (15/2)(12 + 96)
S = (15/2) × 108 = 15 × 54
∴ Sum = 810

Example 6: The 4th term of an AP is 17 and the 9th term is 37. Find the 20th term.

a + 3d = 17 ... (1)
a + 8d = 37 ... (2)
(2) − (1): 5d = 20, so d = 4
a = 17 − 12 = 5
a₂₀ = 5 + 19 × 4
∴ a₂₀ = 81

Example 7: How many terms of 2, 5, 8, ... must be added to get 155?

a = 2, d = 3, Sₙ = 155
(n/2)[2 × 2 + (n − 1) × 3] = 155
(n/2)(3n + 1) = 155
3n² + n − 310 = 0
D = 1 + 4 × 3 × 310 = 3721 = 61²
n = (−1 + 61)/6 = 10 (negative root rejected)
∴ 10 terms

Common Mistakes to Avoid

• Finding d as a₁ − a₂ instead of a₂ − a₁; for a decreasing AP the common difference must come out negative.
• Using n instead of (n − 1) in the nth term formula; the first term has zero steps of d added to it.
• Mixing up aₙ (a single term) and Sₙ (a total); read the question twice to see which one is asked.
• Accepting a fractional or negative value of n; the number of terms must be a positive whole number.

Quick Revision

d = a₂ − a₁
aₙ = a + (n − 1)d
Sₙ = (n/2)[2a + (n − 1)d] = (n/2)(a + l)
aₙ = Sₙ − Sₙ₋₁
a, b, c in AP ⇔ 2b = a + c

• The number of terms n must always be a positive whole number

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