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Arithmetic Progressions — Class 10 Maths Notes & Practice

July 31, 2026
Arithmetic Progressions — Class 10 Maths Notes & Practice

An Arithmetic Progression is simply a list of numbers where you add the same amount each time — like 3, 7, 11, 15. Once you know two small formulas, this becomes one of the quickest chapters to score in. This page teaches it in plain language, with a plan and an original practice set (with answers you can reveal) at the end.

What This Chapter Covers

Your Game Plan for This Chapter

  1. First — Understand the common difference and how to spot an AP.
  2. Next — Learn the n-th term and sum formulas — they solve almost everything.
  3. Last — Practise word problems, then attempt the practice set below.

Study Notes

1. What Is an Arithmetic Progression?

An AP is a sequence in which each term is obtained by adding a fixed number, called the common difference (d), to the previous term. Its general form is a, a + d, a + 2d, a + 3d, …, where a is the first term. You can always find d by subtracting any term from the one after it: $d = a_2-a_1$.

Number line showing the arithmetic progression 3, 7, 11, 15, 19 with a curved arrow labelled plus 4 between each pair of terms, first term a = 3 and common difference d = 4.
Figure: the AP 3, 7, 11, 15, 19 on a number line, adding d = 4 each time · चित्र: संख्या रेखा पर AP और सार्व अंतर d = 4
Key Idea
A sequence is an AP only if the difference between consecutive terms is the same throughout. Check two or three gaps — if they match, it is an AP.

2. The n-th Term

To jump straight to any term without listing them all, use the n-th term formula. For example, in 3, 7, 11, 15, … (a = 3, d = 4), the 15th term is $a_{15} = 3 + (15-1) \times 4 = 3 + 56 = 59$.

Flowchart with five steps for finding the 15th term of an arithmetic progression: given a = 3, d = 4, n = 15; formula an = a + (n minus 1) d; substitute 3 + (15 minus 1) times 4; simplify 3 + 56; answer 15th term = 59.
Figure: the five steps that take a = 3, d = 4, n = 15 to the answer 59 · चित्र: 15वाँ पद निकालने के पाँच चरण

3. Sum of the First n Terms

To add up the first n terms, use the sum formula. These two formulas cover the whole chapter:

FormulaWhat it finds
$a_n = a + (n-1)d$The n-th term of the AP
$S_n = \frac{n}{2}\left[2a + (n-1)d\right]$The sum of the first n terms
$S_n = \frac{n}{2}(a + l)$The sum when the last term l is known
Common Mistake
It is (n − 1)d, not n·d. Forgetting the “minus one” is the single most common error in this chapter — the 1st term uses (1 − 1)d = 0, so $a_1 = a$.

4. Solving Problems and Word Problems

Most questions give you some information and ask for a term, the number of terms, or a sum. Write down a, d and what is asked, choose the right formula, and solve. Word problems — savings that grow by a fixed amount, seats increasing row by row — are just APs in disguise; find a and d, then apply the same formulas.

Exam Tip
Almost every AP question reduces to “find a and d, then use a formula.” Get those two values first, and the rest is a one-line calculation.

Practice Worksheet

Try each question fully on your own first, then click Show Answer to check yourself.

Q1. Find the 15th term of the AP: 3, 7, 11, 15, …

Show Answer
a = 3 and d = 4. $a_{15} = a + (15-1)d = 3 + 14 \times 4 = 3 + 56 = 59$.

Q2. Which term of the AP 5, 8, 11, … is 68?

Show Answer
a = 5, d = 3. Set $a_n = 68$: 5 + (n − 1)3 = 68 → (n − 1)3 = 63 → n − 1 = 21 → n = 22. So 68 is the 22nd term.

Q3. Find the sum of the first 20 terms of the AP: 2, 5, 8, …

Show Answer
a = 2, d = 3, n = 20. $S_{20} = \frac{20}{2}\left[2(2) + (20-1)3\right] = 10\left[4 + 57\right] = 10 \times 61 = 610$.

Q4. The first term of an AP is 7 and its 13th term is 43. Find the common difference.

Show Answer
$a_{13} = a + 12d = 43$, and a = 7, so 7 + 12d = 43 → 12d = 36 → d = 3.

Q5. A person saves ₹100 in the first month and increases the saving by ₹50 each month. How much is saved in the 10th month, and in total over 10 months?

Show Answer
This is an AP with a = 100, d = 50. 10th month: $a_{10} = 100 + 9 \times 50$ = ₹550. Total: $S_{10} = \frac{10}{2}\left[2(100) + 9 \times 50\right] = 5\left[200 + 450\right] = 5 \times 650$ = ₹3,250.

Once these feel easy, you have genuinely finished this chapter. Do not aim for perfect on the first try — aim for one more correct question than yesterday.

Written & reviewed by Team Principal Saab — Meet the team →