Sum of n Terms of an AP: Formula, Examples & How to Find It (Class 10)

In one line: the sum of the first n terms of an AP is Sₙ = n/2 × [2a + (n − 1)d], where a is the first term, d is the common difference and n is how many terms you are adding.

If you also know the last term l, there is a shorter version: Sₙ = n/2 × (a + l). Both give the same answer, because the last term is just l = a + (n − 1)d. So, for the AP 8, 3, −2, …, the sum of the first 22 terms is S₂₂ = 22/2 × [2×8 + (22 − 1)×(−5)] = 11 × (16 − 105) = 11 × (−89) = −979.

Sₙ means “the sum of the first n terms.” The clever idea behind the formula (the one Gauss used as a schoolboy) is to add the list to its own reverse: pair the first term with the last, the second with the second-last, and so on. Every pair adds up to the same total, (a + l), and there are n terms, giving n/2 pairs — which is exactly Sₙ = n/2(a + l). The rest of this guide shows you how to pick the right formula, add up an AP, find how many terms reach a target sum, and pull a single term back out of Sₙ.

🎬 Watch (60-sec Short): Can you find the sum of the first 20 terms? ⚡

This guide assumes you already know the nth term formula. If aₙ = a + (n − 1)d is new, read nth Term of an AP first, then come back — the sum formula builds straight on it.

The sum of n terms of an AP formula

For an AP with first term a and common difference d, the sum of the first n terms is:

Sₙ = n/2 × [2a + (n − 1)d]

What it means in plain words: to add the first n terms, take half of n and multiply it by “twice the first term plus (n − 1) times the common difference.”

What each symbol means:

• Sₙ — the sum of the first n terms (the “answer”).
• a — the first term of the AP.
• d — the common difference (subtract any term from the next: d = a₍ₖ₊₁₎ − aₖ).
• n — the number of terms being added (a positive whole number).

When it applies: to any AP, whenever you know a, d and n. If you know the last term l instead of d, use the partner formula below.

Worked example. Find the sum of the first 22 terms of the AP 8, 3, −2, …

• a = 8, d = 3 − 8 = −5, n = 22.
• Sₙ = n/2 × [2a + (n − 1)d] → S₂₂ = 22/2 × [2×8 + (22 − 1)(−5)]
• = 11 × [16 − 105] = 11 × (−89) = −979.

The sum is −979 — negative, because most terms in this decreasing AP are negative. A tidy habit that examfront’s Mistake Identification reinforces: simplify the inside bracket fully before multiplying by n/2, so a stray sign can’t slip through.

The Sₙ = n/2(a + l) formula (when the last term is known)

When a question gives you the first term a and the last term l (but not the common difference), use this shorter form:

Sₙ = n/2 × (a + l)

Here l is the last term of the (finite) AP. This is the direct result of Gauss’s pairing trick, and it is often faster because there is no d to handle.

Worked example. The first and last terms of an AP are 17 and 350, and it has 20 terms. Find the sum.

• a = 17, l = 350, n = 20.
• Sₙ = n/2 × (a + l) → S₂₀ = 20/2 × (17 + 350) = 10 × 367 = 3670.

A classic use — adding consecutive integers. The sum of the first 1000 positive integers is S₁₀₀₀ = 1000/2 × (1 + 1000) = 500 × 1001 = 500500. More generally, the sum of the first n positive integers is Sₙ = n(n + 1)/2 — a result worth memorising. Which form to reach for is a judgement you sharpen quickly with Topic Practice on examfront: last term given → use n/2(a + l); common difference given → use n/2[2a + (n − 1)d].

How to find how many terms give a required sum

Here the sum Sₙ is known and you want n. Because n appears twice in Sₙ = n/2[2a + (n − 1)d], this becomes a quadratic equation in n.

Method: put the target sum into the formula, expand into a quadratic, solve for n, and keep only positive whole-number answers.

Worked example. How many terms of the AP 24, 21, 18, … must be taken so that their sum is 78?

• a = 24, d = −3, Sₙ = 78.
• 78 = n/2 × [48 + (n − 1)(−3)] = n/2 × (51 − 3n)
• 156 = 51n − 3n² → 3n² − 51n + 156 = 0 → n² − 17n + 52 = 0
• (n − 4)(n − 13) = 0 → n = 4 or n = 13.

Both answers are correct — the sum of the first 4 terms and the sum of the first 13 terms are each 78. This surprising “two answers” case happens because the AP starts positive but d is negative, so terms 5 through 13 add up to zero and cancel out. Always keep every positive-integer solution; don’t throw one away.

🎬 Watch (60-sec Short): Can you spot why this AP has two valid answers? 🤯

Watch out: if solving gives a negative or fractional n, reject only that value. If it gives two valid whole numbers, both are genuine answers.

Finding a missing quantity (a, d, n or Sₙ)

The formula links four quantities — Sₙ, a, d and n. Know any three and you can find the fourth. These “fill in the missing piece” questions are common in exams.

Worked example. The sum of the first 14 terms of an AP is 1050 and its first term is 10. Find the 20th term.

• S₁₄ = 1050, n = 14, a = 10. Use Sₙ = n/2[2a + (n − 1)d]:
• 1050 = 14/2 × [20 + 13d] = 7 × (20 + 13d) = 140 + 91d
• 910 = 91d → d = 10.
• Now the 20th term: a₂₀ = a + (20 − 1)d = 10 + 19×10 = 200.

Notice how the sum formula and the nth-term formula work together in one problem — a very common exam pattern. Mixed questions like this are exactly where Chapter Quizzes and personalised help on examfront show you which step trips you up.

The link between Sₙ and the nth term

There is a neat relationship connecting sums and terms:

aₙ = Sₙ − Sₙ₋₁

In plain words: the nth term equals the sum of the first n terms minus the sum of the first (n − 1) terms — because subtracting those two sums leaves only the nth term. This lets you recover a single term when a question gives you a formula for Sₙ.

Worked example. If the sum of the first n terms of an AP is Sₙ = 2n² + 3n, find the 10th term.

• S₁₀ = 2(10)² + 3(10) = 200 + 30 = 230.
• S₉ = 2(9)² + 3(9) = 162 + 27 = 189.
• a₁₀ = S₁₀ − S₉ = 230 − 189 = 41.

🎬 Watch (60-sec Short): Can you find the 10th term from the sum formula? 🔎

Where this leads next

You now have the two big tools of Chapter 5 working together: the nth term to reach any single term, and the sum Sₙ to add many terms at once. Together they solve almost every AP word problem — total savings, stacked logs, prize money, distances covered, and more.

We have kept this guide to the core formulas and their most common question types. The trickier applications — spiral and staircase problems, AP word problems with hidden data, questions that combine Sₙ with the nth term in several steps, and full timed practice sets — are where you push for top marks. Work through the complete graded set inside examfront, and let Progress Tracking and Strength & Weakness Analysis show you exactly which AP-sum skills are exam-ready. For the full chapter map, see the Arithmetic Progressions chapter guide.

Key Takeaways

• The sum of the first n terms of an AP is Sₙ = n/2[2a + (n − 1)d].
• If the last term l is known, use the shorter Sₙ = n/2(a + l) — both always agree.
• The sum of the first n positive integers is Sₙ = n(n + 1)/2.
• “How many terms give this sum?” leads to a quadratic in n — keep every positive-integer answer (there can be two).
• The nth term and the sum are linked by aₙ = Sₙ − Sₙ₋₁.

Quick Facts

• Sum formula (d known): Sₙ = n/2[2a + (n − 1)d].
• Sum formula (last term known): Sₙ = n/2(a + l).
• First n positive integers: Sₙ = n(n + 1)/2.
• Sₙ: sum of first n terms. a: first term. d: common difference. l: last term. n: number of terms.
• Term from sum: aₙ = Sₙ − Sₙ₋₁.
• Target-sum questions: solve a quadratic in n; two positive-integer answers can both be valid.
• Chapter: Arithmetic Progressions (Ch. 5) · Class: 10 · Subject: Maths · Board: CBSE.

Common Mistakes

1. Using n as the last term. Sₙ = n/2(a + l) needs the last term l, not n. In “sum of 5, 8, …, 50,” the last term is 50, so use a = 5, l = 50 — don’t plug in n where l belongs.
2. Forgetting the ½ (the n/2). Writing Sₙ = n[2a + (n − 1)d] doubles the answer. The formula is half of n times the bracket — keep the /2.
3. Discarding a valid second value of n. In target-sum questions the quadratic can give two positive whole numbers (like n = 4 and n = 13), and both are correct. Reject only negative or fractional n.
4. Sign slips inside the bracket. With a negative d, terms like (n − 1)(−3) must keep their minus sign. Simplify 2a + (n − 1)d completely before multiplying by n/2.
5. Mixing up Sₙ and aₙ. Sₙ is the total of many terms; aₙ is one single term. If a question asks for “the 20th term,” don’t stop at the sum — use aₙ = a + (n − 1)d, or aₙ = Sₙ − Sₙ₋₁.

FAQ

Q. What is the sum of n terms of an AP? The sum of the first n terms, written Sₙ, is Sₙ = n/2[2a + (n − 1)d], where a is the first term, d is the common difference and n is the number of terms. If you know the last term l instead of d, use Sₙ = n/2(a + l). For example, the sum of the first 22 terms of 8, 3, −2, … is −979.

Q. What is the formula for the sum of an arithmetic progression? There are two equivalent formulas: Sₙ = n/2[2a + (n − 1)d] when you know a and d, and Sₙ = n/2(a + l) when you know the first term a and last term l. Both give the sum of the first n terms — choose the one matching the information you have.

Q. When do I use Sₙ = n/2(a + l) instead of Sₙ = n/2[2a + (n − 1)d]? Use Sₙ = n/2(a + l) when the last term l is given (for example, “the first and last terms are 17 and 350”). Use Sₙ = n/2[2a + (n − 1)d] when you know d but not the last term. They always agree, because l = a + (n − 1)d.

Q. How do I find how many terms of an AP give a certain sum? Put the target sum into Sₙ = n/2[2a + (n − 1)d] and solve the resulting quadratic for n, keeping only positive whole-number values. Sometimes two values of n give the same sum — this happens when positive and negative terms cancel, so both answers are correct.

Q. How are Sₙ and the nth term related? The nth term equals the sum to n terms minus the sum to (n − 1) terms: aₙ = Sₙ − Sₙ₋₁. So if you are given a formula for Sₙ, you can find any single term by subtracting consecutive sums. For Sₙ = 2n² + 3n, a₁₀ = S₁₀ − S₉ = 230 − 189 = 41.

Related Concepts

• nth Term of an AP (Class 10): the aₙ = a + (n − 1)d formula this sum builds on, and the last term l used in Sₙ = n/2(a + l).
• What Is an Arithmetic Progression? (Class 10): the first term, common difference and general form — the foundation for both formulas.
• Arithmetic Progressions — Chapter Guide (Class 10): the full chapter roadmap and where the sum formula fits.
• Class 10 Maths — All Formulas: the quick-reference sheet with both AP-sum formulas.

Ready to make AP sums automatic? Practise a full graded set on examfront’s Topic Practice, and let Mistake Identification and Strength & Weakness Analysis show you exactly where your sign slips and dropped-½ errors happen. Start on examfront →