Arithmetic Progressions Class 10: Notes and Important Questions

This guide covers Arithmetic Progressions Class 10, chapter 5 of NCERT Class 10 Mathematics. You will learn what makes a list of numbers an AP, how to find any term with the nth term formula, and how to add up the first n terms quickly. Solved examples, important questions and common mistakes are all here.

Patterns that grow by a fixed amount

Many lists of numbers grow or shrink by the same amount each time.

• A monthly salary that starts at 8000 and rises by 500 every year: 8000, 8500, 9000, ...
• Ladder rungs that start at 45 cm and get 2 cm shorter each time: 45, 43, 41, 39, ...
• 100, 150, 200, 250, ...: each term is 50 more than the one before.

In every case, each term after the first is obtained by adding a fixed number to the previous term. That is the pattern this chapter studies.

What an AP is

An arithmetic progression (AP) is a list of numbers in which each term, except the first, is obtained by adding a fixed number to the preceding term.

• The first term is called a.
• The fixed number is the common difference, d.

d can be positive (terms increase), negative (terms decrease) or zero (all terms equal, like 3, 3, 3, 3, ...).

To test a list, find a₂ − a₁, a₃ − a₂, a₄ − a₃, ... If all these differences are the same, the list is an AP.

A finite AP has a last term. An infinite AP goes on forever.

General form: a, a + d, a + 2d, a + 3d, ... For example, a = 6 and d = 3 give 6, 9, 12, 15, ...

Example: for 3/2, 1/2, −1/2, −3/2, ..., a = 3/2 and d = 1/2 − 3/2 = −1.

The nth term

Look at the pattern: a₂ = a + d, a₃ = a + 2d, a₄ = a + 3d. The number of d's is always one less than the position. So

aₙ = a + (n − 1)d

Here n is the position of the term, and it must be a positive integer. The nth term is also called the general term. The last term of a finite AP is often written l.

Example: find the 10th term of 2, 7, 12, ... Here a = 2, d = 5, n = 10. a₁₀ = 2 + 9 × 5 = 47.

Is a number a term of the AP? Solve for n. If n comes out as 1, 2, 3, ..., it is a term. If n is zero, negative or a fraction, it is not.

Using the nth term

Counting terms. How many two-digit numbers are divisible by 3? The first is 12, the last is 99 and d = 3. So 99 = 12 + (n − 1)3, which gives 87 = 3(n − 1), n − 1 = 29 and n = 30.

Term from the end. Find the 11th term from the last term of 10, 7, 4, ..., −62. Reverse the AP: it starts at −62 with d = +3. Its 11th term is −62 + 10 × 3 = −32.

Sum of the first n terms

The sum formula comes from the method used by the mathematician Gauss: write the sum forwards and backwards, then add the two lines.

Sₙ = (n/2)2a + (n − 1)d

If the last term l is known:

Sₙ = (n/2)(a + l)

The formula links four quantities: Sₙ, n, a and d. If you know any three, you can find the fourth.

Also, a₁ = S₁ and, for n > 1, aₙ = Sₙ − Sₙ₋₁. Subtracting the sum of the first n − 1 terms leaves only the nth term. Do not confuse aₙ (one term) with Sₙ (a total).

Example: a = 100, d = 50, n = 21. S₂₁ = (21/2)200 + 20 × 50 = (21/2)(1200) = 21 × 600 = 12600.

Example: the sum of the first 1000 positive integers is (1000/2)(1 + 1000) = 500 × 1001 = 500500.

Harder sum problems

When n comes from a quadratic. For 24, 21, 18, ..., how many terms give a sum of 78? Using the sum formula leads to n² − 17n + 52 = 0, so n = 4 or n = 13. Both are correct. Because d is negative, later terms become negative and cancel some of the positive ones, so two different n give the same sum. Check: 24 + 21 + 18 + 15 = 78.

When aₙ is given as an expression. If aₙ = 3 + 2n, then a₁ = 5 and a₂ = 7, so a = 5 and d = 2. S₂₄ = 1210 + 23 × 2 = 12 × 56 = 672.

Arithmetic mean

If a, b, c are in AP, then b − a = c − b, so 2b = a + c and

b = (a + c)/2

b is called the arithmetic mean of a and c. For 2, , 26 the missing term is (2 + 26)/2 = 14. For 5, x, 15, x = 10.

Remember this

• AP: equal differences between consecutive terms.
• aₙ = a + (n − 1)d, with n a positive integer.
• Sₙ = (n/2)2a + (n − 1)d = (n/2)(a + l).
• a₁ = S₁ and aₙ = Sₙ − Sₙ₋₁.
• Middle of three terms in AP: b = (a + c)/2.

Important questions with answers

1. Can the common difference of an AP be zero? Yes. Then every term is the same, for example 5, 5, 5, 5, ...

2. Write the first four terms of the AP with a = −2 and d = 0. −2, −2, −2, −2.

3. If a = 21 and d = −3, which term is −81? −81 = 21 + (n − 1)(−3), so −102 = −3(n − 1), n − 1 = 34 and n = 35. It is the 35th term.

4. How many terms are in 7, 13, 19, ..., 205? 205 = 7 + (n − 1)6, so 198 = 6(n − 1), n − 1 = 33 and n = 34.

5. Find the 11th term from the end of 10, 7, 4, ..., −62. Reversed AP: a = −62, d = 3. 11th term = −62 + 10 × 3 = −32.

6. Find S₂₄ if aₙ = 3 + 2n. a = 5, d = 2. S₂₄ = 12(10 + 46) = 672.

7. If Sₙ = 4n − n², find the first term and the second term. a₁ = S₁ = 4 − 1 = 3. S₂ = 8 − 4 = 4, so a₂ = S₂ − S₁ = 4 − 3 = 1. (So d = −2.)

8. If 10, x, 30 are in AP, find x. x = (10 + 30)/2 = 20.

9. How many terms of 24, 21, 18, ... add up to 78? n = 4 or n = 13; both work because d is negative.

Common mistakes to avoid

• Writing aₙ = a + nd. The coefficient of d is n − 1.
• Finding d as a₁ − a₂ instead of a₂ − a₁, which flips the sign.
• Accepting a fractional or negative n as a term position.
• Forgetting that counting from the end reverses the sign of d.
• Mixing up aₙ (one term) and Sₙ (the total).
• Throwing away one of two valid values of n without checking.

Write out the first few terms whenever you feel unsure, and study this chapter with Joy to practise step by step.