Class 10 Maths Arithmetic Progressions Notes

Arithmetic Progressions – Class 10 Maths Notes

1. Introduction

In mathematics and everyday life, we often come across number patterns. For example, salaries may increase by a fixed amount every year, savings may increase by a fixed amount every month, or the number of objects in successive rows may decrease by a fixed amount.

A sequence in which each term is obtained by adding the same fixed number to the preceding term is called an Arithmetic Progression, or AP.

2. What is an Arithmetic Progression?

An Arithmetic Progression is a list of numbers in which the difference between consecutive terms remains constant.

For example:

2, 5, 8, 11, 14, …

Here:

5 − 2 = 3
8 − 5 = 3
11 − 8 = 3

Since the difference is always 3, the sequence is an AP.

The general form of an AP is:

a, a + d, a + 2d, a + 3d, …

where:

a = first term
d = common difference

The common difference may be positive, negative, or zero.

Examples of APs

1, 2, 3, 4, 5, …
- First term = 1
- Common difference = 1
20, 17, 14, 11, …
- First term = 20
- Common difference = −3
5, 5, 5, 5, …
- First term = 5
- Common difference = 0
−2, −1, 0, 1, 2, …
- First term = −2
- Common difference = 1

3. How to Identify an AP

To check whether a sequence is an AP, subtract each term from the term immediately after it.

The differences must be equal.

For example:

6, 10, 14, 18, …

10 − 6 = 4
14 − 10 = 4
18 − 14 = 4

Therefore, it is an AP with common difference 4.

But consider:

1, 2, 4, 8, …

2 − 1 = 1
4 − 2 = 2
8 − 4 = 4

The differences are not equal.

Therefore, it is NOT an AP.

Formula for common difference

d = next term − previous term

For an AP:

d = a₂ − a₁ = a₃ − a₂ = a₄ − a₃ = …

4. First Term and Common Difference

The first term of an AP is generally represented by a.

The common difference is represented by d.

For example:

7, 10, 13, 16, …

Here:

a = 7

d = 10 − 7 = 3

Therefore, the AP can be written as:

7, 7 + 3, 7 + 2(3), 7 + 3(3), …

or simply:

7, 10, 13, 16, …

Knowing the first term and common difference is enough to construct the entire AP.

5. General Term or nth Term of an AP

Consider the AP:

a, a + d, a + 2d, a + 3d, …

The terms can be written as:

First term = a

Second term = a + d

Third term = a + 2d

Fourth term = a + 3d

Therefore, the nth term is:

nth term = a + (n − 1)d

In symbols:

aₙ = a + (n − 1)d

This is one of the most important formulas in the chapter.

6. Finding a Particular Term

Example:

Find the 10th term of:

2, 7, 12, 17, …

Here:

a = 2

d = 7 − 2 = 5

n = 10

Using:

aₙ = a + (n − 1)d

a₁₀ = 2 + (10 − 1)(5)

a₁₀ = 2 + 45

a₁₀ = 47

Therefore:

10th term = 47

7. Finding Which Term Contains a Given Number

Sometimes we are given a number and asked whether it occurs in an AP.

Use:

aₙ = a + (n − 1)d

Then substitute the given number for aₙ and solve for n.

If n is a positive integer, the number is a term of the AP.

If n is not a positive integer, the number is not a term.

Example

Check whether 301 is a term of:

5, 11, 17, 23, …

Here:

a = 5

d = 6

Suppose 301 is the nth term.

301 = 5 + (n − 1)6

301 = 5 + 6n − 6

301 = 6n − 1

302 = 6n

n = 302/6 = 151/3

Since n is not an integer:

301 is not a term of the AP.

8. Finding an AP When Two Terms Are Given

If two terms of an AP are known, use the nth-term formula to form two equations.

Example

The 3rd term of an AP is 5 and the 7th term is 9. Find the AP.

For the 3rd term:

a₃ = a + (3 − 1)d

5 = a + 2d

For the 7th term:

a₇ = a + (7 − 1)d

9 = a + 6d

So:

a + 2d = 5
a + 6d = 9

Subtracting:

4d = 4

d = 1

Substitute into the first equation:

a + 2 = 5

a = 3

Therefore, the AP is:

3, 4, 5, 6, 7, …

9. Finding a Term from the End

Suppose an AP has n terms.

The kth term from the end is:

Term from the end = (n − k + 1)th term from the beginning

For example, if an AP has 25 terms, then:

11th term from the end = (25 − 11 + 1)th term

= 15th term from the beginning.

This method is useful in questions involving the last term of an AP.

10. Sum of the First n Terms of an AP

Suppose the AP is:

a, a + d, a + 2d, …, a + (n − 1)d

The sum of the first n terms is denoted by Sₙ.

The formula is:

Sₙ = n/2 [2a + (n − 1)d]

This is the main formula for finding the sum of an AP.

11. Sum Formula Using the Last Term

If the last term of an AP is l, then:

Sₙ = n/2 (a + l)

where:

Sₙ = sum of n terms
n = number of terms
a = first term
l = last term

This formula is especially useful when the first and last terms are given directly.

Easy way to remember

Sum of AP = Number of terms × Average of first and last terms

Therefore:

Sₙ = n × (a + l)/2

12. Example of Sum of an AP

Find the sum of the first 22 terms of:

8, 3, −2, …

Here:

a = 8

d = 3 − 8 = −5

n = 22

Using:

Sₙ = n/2 [2a + (n − 1)d]

S₂₂ = 22/2 [2(8) + (22 − 1)(−5)]

S₂₂ = 11 [16 − 105]

S₂₂ = 11(−89)

S₂₂ = −979

Therefore:

Sum = −979

13. Relationship Between nth Term and Sum

There is an important relationship:

aₙ = Sₙ − Sₙ₋₁

This means that the nth term can be obtained by subtracting the sum of the first (n − 1) terms from the sum of the first n terms.

14. Sum of the First n Positive Integers

The sequence:

1, 2, 3, 4, …, n

is an AP.

Here:

a = 1

d = 1

l = n

Using:

Sₙ = n/2 (a + l)

we get:

Sₙ = n(n + 1)/2

Therefore:

1 + 2 + 3 + … + n = n(n + 1)/2

Example:

1 + 2 + 3 + … + 100

= 100(101)/2

= 5050

15. Arithmetic Mean

If three numbers a, b and c are in AP, then b is called the arithmetic mean of a and c.

Therefore:

b = (a + c)/2

or:

2b = a + c

Example

Find the number that must be placed between 8 and 20 so that the three numbers are in AP.

Let the middle number be x.

Then:

2x = 8 + 20

2x = 28

x = 14

Therefore:

8, 14, 20 is an AP.

The arithmetic mean is 14.

16. APs in Real-Life Situations

Arithmetic progressions are useful when a quantity changes by a fixed amount repeatedly.

Example 1: Salary

Suppose a person’s starting monthly salary is Rs. 8,000 and it increases by Rs. 500 every year.

The salaries form:

8,000, 8,500, 9,000, 9,500, …

Here:

a = 8000

d = 500

Salary in the nth year:

aₙ = 8000 + (n − 1)(500)

For example, salary in the 25th year:

a₂₅ = 8000 + 24(500)

= Rs. 20,000

The chapter uses this type of situation to introduce the nth-term formula.

17. AP and Simple Interest

In a simple-interest situation, if the principal and rate remain fixed, the interest earned at the end of successive years increases by a fixed amount.

For example, for Rs. 1,000 at 8% simple interest:

1st year interest = Rs. 80
2nd year interest = Rs. 160
3rd year interest = Rs. 240
…

This forms an AP:

80, 160, 240, …

Here:

a = 80

d = 80

Therefore, interest after 30 years is:

a₃₀ = 80 + 29(80)

= Rs. 2,400.

18. How to Solve Word Problems on AP

When a word problem gives a sequence that increases or decreases uniformly, follow these steps:

Step 1: Write the sequence

Example:

23, 21, 19, …, 5

Step 2: Identify a and d

a = 23

d = 21 − 23 = −2

Step 3: Identify what is required

If the last term is known and the number of terms is required, use:

aₙ = a + (n − 1)d

Step 4: Substitute the values

5 = 23 + (n − 1)(−2)

5 = 23 − 2n + 2

5 = 25 − 2n

2n = 20

n = 10

Therefore:

There are 10 terms.

19. Important Formula Sheet