Arithmetic Progressions — nth Term and Sum Made Simple

Understand Arithmetic Progressions, the common difference, and the formulas for the nth term and the sum of n terms, explained with a staircase analogy, a fully worked example and a quick self-check quiz.

By the PadhoDost Team · 📖 7 min read · Updated 4 August 2026

Part of Class 10 (CBSE) prep

🧠 Climbing an Even Staircase

Picture a staircase where every step rises by exactly the same height, say 15 cm. The 1st step is 15 cm above the floor, the 2nd is 30 cm, the 3rd is 45 cm, and so on. Each number is simply the one before it plus a fixed jump. That steady, equal jump is the whole idea behind an Arithmetic Progression — a list of numbers that grows or shrinks by the same amount every single time.

What Makes a Sequence an AP?

37111519 +4+4+4+4 aₙ = a + (n − 1)d
In an AP each term jumps by the same common difference d — here d = 4.

An Arithmetic Progression (AP) is a list of numbers in which the difference between any term and the term just before it is always the same. This fixed difference is called the common difference, written as d, and the first term is called a. To test whether a sequence is an AP, subtract each term from the next — if you always get the same number, it is an AP.

General form: a, a + d, a + 2d, a + 3d, ... where a = first term and d = common difference
nth term: a_n = a + (n - 1)d
💡 Find the common difference by subtracting any term from the one right after it: d = a_2 - a_1 = a_3 - a_2. If these differences are not all equal, the sequence is NOT an AP.

Adding Up the Terms

Sum of first n terms: S_n = (n/2) x [ 2a + (n - 1)d ] or S_n = (n/2) x (a + l), where l is the last term

📝 Worked example: the 15th term and a sum

Consider the AP: 3, 7, 11, 15, ...

First term a = 3. Common difference d = 7 - 3 = 4.

Find the 15th term using a_n = a + (n - 1)d:

a_15 = 3 + (15 - 1) x 4 = 3 + 14 x 4 = 3 + 56 = 59

Now find the sum of the first 20 terms using S_n = (n/2)[2a + (n - 1)d]:

S_20 = (20/2) x [2 x 3 + (20 - 1) x 4] = 10 x [6 + 76] = 10 x 82 = 820

So the 15th term is 59 and the sum of the first 20 terms is 820.

QuantityFormula
Common differenced = a_2 - a_1
nth terma_n = a + (n - 1)d
Sum of n termsS_n = (n/2)[2a + (n - 1)d]
Sum using last term lS_n = (n/2)(a + l)

Remember these

  • An AP has a constant common difference d between consecutive terms.
  • nth term: a_n = a + (n - 1)d — use it to jump straight to any term.
  • Sum of n terms: S_n = (n/2)[2a + (n - 1)d].
  • If the last term l is known, S_n = (n/2)(a + l).
  • d can be positive (increasing AP), negative (decreasing AP), or zero (all terms equal).

⚡ Quick check

Which term of the AP 5, 8, 11, 14, ... is equal to 50?

Ready to test yourself? 🎯

Lock it in with the practice test for this chapter.

Take the practice test →

Keep studying

See all Class 10 (CBSE) study material →