Class 10 Maths Arithmetic Progressions Notes by PW

An Arithmetic Progression (A.P.) is an important topic in Class 10 Maths. It connects sequences with linear equations, inequalities, divisibility, ratios and sums. You may find questions asking you to identify a missing term, calculate the number of terms or find the sum of a particular progression.

Physics Wallah offers Arithmetic Progressions notes to help you understand these concepts and apply the correct formulas. Learning how to identify the first term, common difference and position of a term makes it easier to solve different types of A.P. questions in your Class 10 Maths board exams.

Arithmetic Progressions

Basics

An Arithmetic Progression (A.P.) is a sequence in which the difference between consecutive terms is constant.

If the terms are a₁, a₂, a₃, ..., then:

a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = d

Here:

• a₁ = first term

• d = common difference

• aₙ = nth term

• n = position of the term

An A.P. can be finite (it has a last term) or infinite (it continues forever).

General form: a₁, a₁ + d, a₁ + 2d, a₁ + 3d, ...

nth term formula:

aₙ = a₁ + (n − 1)d

Finding the Common Difference

Subtract any term from the term that follows it:

d = aₙ₊₁ − aₙ

Example: 7, 12, 17, 22, ...
12 − 7 = 5, 17 − 12 = 5, 22 − 17 = 5, so d = 5.

Behaviour of Terms

• If d > 0, the terms increase.

• If d < 0, the terms decrease.

• If d = 0, all terms are equal.

The common difference can be a fraction or negative.

• a₁ = 5/2, d = 1/2 gives 5/2, 3, 7/2, 4, ...

• a₁ = 4, d = −3 gives 4, 1, −2, −5, ...

Finding the nth Term

The formula lets you find any term without listing all the earlier ones.

Example: 9, 13, 17, ...
a₁ = 9, d = 13 − 9 = 4
aₙ = 9 + (n − 1) × 4 = 9 + 4n − 4

aₙ = 4n + 5

Check: a₁ = 9, a₂ = 13, a₃ = 17, a₄ = 21. The formula works.

Finding a Term from the End

Reverse the A.P. The last term becomes the first term, and d changes sign.

Memory Tip: To find a term from the end, reverse the A.P. and change the sign of d.

Example: An A.P. ends at 257 and has d = 4. Find the 10th term from the end.
For the reversed A.P., a₁ = 257 and d = −4.
a₁₀ = 257 + (10 − 1)(−4) = 257 − 36

a₁₀ = 221

Finding Which Term Has a Given Value

Put the value equal to aₙ and solve for n:

a₁ + (n − 1)d = given value

The number belongs to the A.P. only if n is a positive integer.

Memory Tip: n cannot be negative or fractional. Always check this before concluding.

Finding an A.P. from Two Conditions

Form two equations using the nth term formula and solve them together.

Example: The 3rd term is 16, and the 7th term exceeds the 5th term by 12.

a₃ = 16, so a₁ + 2d = 16 ...(1)
a₇ − a₅ = 12, so (a₁ + 6d) − (a₁ + 4d) = 12
2d = 12, so d = 6

From (1): a₁ + 12 = 16, so a₁ = 4.

Required A.P.: 4, 10, 16, 22, 28, ...

Finding Terms When Two Terms Are Given

Form one equation for each given term.

Example: a₇ = 1/9 and a₉ = 1/7. Find a₆₃.

a₁ + 6d = 1/9 ...(1)
a₁ + 8d = 1/7 ...(2)

Subtracting (1) from (2):
2d = 1/7 − 1/9 = 2/63

d = 1/63

Substituting in (1):
a₁ = 1/9 − 6/63 = 7/63 − 6/63

a₁ = 1/63

a₆₃ = a₁ + 62d = 1/63 + 62/63

a₆₃ = 1

Multiples as an A.P.

Multiples of any fixed positive number form an A.P. For example, the multiples of 7 are 7, 14, 21, 28, ..., with a₁ = 7 and d = 7.

Counting three-digit multiples of 7

The first and last three-digit multiples of 7 are 105 and 994.
a₁ = 105, aₙ = 994, d = 7

994 = 105 + (n − 1) × 7
889 = 7(n − 1)
n − 1 = 127

n = 128

There are 128 three-digit numbers divisible by 7.

Consecutive Terms in an A.P.

If three expressions T₁, T₂, T₃ are consecutive terms of an A.P., then:

T₂ − T₁ = T₃ − T₂

Substitute the given expressions, equate the two differences and solve for the unknown (such as k).

Finding the First Negative Term

Set aₙ < 0 and solve the inequality.

Example: a₁ = 65, d = −4

65 + (n − 1)(−4) < 0
65 − 4n + 4 < 0
69 < 4n
n > 17.25

Since n must be a positive integer, the smallest value is n = 18.

a₁₈ = 65 + 17(−4) = 65 − 68 = −3

So the 18th term is the first negative term.

Memory Tip: Set aₙ < 0, solve the inequality and pick the smallest positive integer that satisfies it.

Middle Term of an A.P.

• If the number of terms n is odd, the middle term is the ((n + 1)/2)th term.

• If n is even, there are two middle terms: the (n/2)th and the (n/2 + 1)th terms.

Examples:

• For 11 terms, (11 + 1)/2 = 6, so the 6th term is the middle term.

• For 10 terms, the middle terms are the 5th and 6th terms.

Sum of the First n Terms

Sₙ denotes the sum of the first n terms.

When a₁ and d are known:

Sₙ = (n/2) [2a₁ + (n − 1)d]

When a₁ and the last term aₙ are known:

Sₙ = (n/2) (a₁ + aₙ)

Choose the formula based on the information given in the question.

Sum of three-digit multiples of 11

The A.P. is 110, 121, 132, ..., 990, with a₁ = 110, aₙ = 990 and d = 11.

Number of terms:
990 = 110 + (n − 1) × 11
880 = 11(n − 1)
n − 1 = 80, so n = 81

Sum:
S₈₁ = (81/2)(110 + 990) = (81/2) × 1100

S₈₁ = 44,550

Number of Terms for a Zero Sum

Put Sₙ = 0 in the sum formula and solve for n.

Example: 27, 24, 21, ... (a₁ = 27, d = −3)

(n/2)[2(27) + (n − 1)(−3)] = 0
(n/2)[54 − 3(n − 1)] = 0
(n/2)(57 − 3n) = 0

Ignore n = 0, since it is not a valid number of terms.
57 − 3n = 0, so 3n = 57

n = 19

The sum of the first 19 terms is zero.

Ratio of Terms and Sums

For ratio questions, substitute the general terms and simplify.

Example: a₁₁ / a₁₈ = 2/3. Find S₅ : S₁₀.

(a₁ + 10d) / (a₁ + 17d) = 2/3
3(a₁ + 10d) = 2(a₁ + 17d)
3a₁ + 30d = 2a₁ + 34d

a₁ = 4d

S₅ = (5/2)[2a₁ + 4d] = (5/2)[8d + 4d] = 30d
S₁₀ = (10/2)[2a₁ + 9d] = 5[8d + 9d] = 85d

S₅ / S₁₀ = 30d / 85d = 6/17

S₅ : S₁₀ = 6 : 17

Sum of Consecutive Groups of Terms

When the sums of two consecutive groups of terms are given, first find the total sum, then form equations using the sum formula.

Example: The sum of the first 7 terms is 63, and the sum of the 8th to 14th terms is 161. Find a₂₈.

S₁₄ = 63 + 161 = 224

From S₇: (7/2)[2a₁ + 6d] = 63
Dividing by 7: a₁ + 3d = 9 ...(1)

From S₁₄: (14/2)[2a₁ + 13d] = 224
2a₁ + 13d = 32 ...(2)

Multiply (1) by 2: 2a₁ + 6d = 18
Subtract this from (2): 7d = 14

d = 2

From (1): a₁ + 6 = 9

a₁ = 3

a₂₈ = a₁ + 27d = 3 + 27(2)

a₂₈ = 57

Arithmetic Progressions are sequences in which the difference between consecutive terms stays constant. The nth term formula, aₙ = a₁ + (n − 1)d, helps you find any term, check whether a number belongs to the A.P. and form equations from given conditions. The sum formulas, Sₙ = (n/2)[2a₁ + (n − 1)d] and Sₙ = (n/2)(a₁ + aₙ), help you solve questions on multiples, zero sums, ratios and groups of terms. Revise both formulas, the first negative term and middle term methods, and practise questions from the NCERT solutions for Arithmetic Progressions to strengthen your preparation for Class 10 Maths.