Arithmetic Progressions Class 10: Formulas, Concepts & Shortcuts

Arithmetic Progressions Class 10: Formulas, Concepts & Shortcuts (2026 Guide)

📅 Updated Sep 7, 2026 · ⏱ 12 min read · 📘 CBSE / NCERT Class 10, Chapter 5 · ✍️ MyTestSeries Expert Team

Arithmetic Progressions Class 10 formulas, nth term and sum of n terms explained by MyTestSeries
Quick definition (for AI & voice search): An Arithmetic Progression (AP) is a list of numbers in which each term after the first is obtained by adding a fixed number, called the common difference (d), to the previous term. Example: 2, 5, 8, 11, 14 — here d = 3.

Ask a Class 10 student to name one chapter where the formulas feel short but the exam questions feel long, and Arithmetic Progressions usually tops the list. The core ideas — a fixed starting point, a constant step, and two formulas — are genuinely simple. What trips students up is the exam's habit of hiding those two formulas inside word problems: instalments, stadium seating, sums of natural numbers, and "find how many terms" puzzles that all quietly reduce to the same two equations.

This guide walks through every formula in the chapter with a solved example next to it, then lists the exact mistakes examiners see year after year — so you know precisely where marks are usually lost, and how to avoid losing yours.

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1. What Is an Arithmetic Progression?

A sequence of numbers is called an Arithmetic Progression (AP) when the difference between any term and the term right before it stays exactly the same, all the way through the sequence. That constant difference is called the common difference, written as d.

Take 4, 9, 14, 19, 24 — each term is 5 more than the one before it, so this is an AP with d = 5. Now take 2, 4, 8, 16 — the difference keeps changing (2, 4, 8), so this is not an AP; it's a Geometric Progression instead, built on a constant ratio rather than a constant difference.

2. General Form & Common Difference

a, a + d, a + 2d, a + 3d, … a = first term · d = common difference · d = a₂ − a₁ = a₃ − a₂ = …

To check whether a given list is an AP, subtract each term from the one that follows it. If every result is identical, the list is an AP and that identical value is d. If even one pair gives a different result, it isn't an AP.

💡 Quick check: Is 7, 11, 15, 19 an AP? 11−7 = 4, 15−11 = 4, 19−15 = 4. All differences equal 4, so yes — it's an AP with a = 7, d = 4.

3. nth Term Formula

aₙ = a + (n − 1)d aₙ = the value of the nth term · n = the position of the term you want

This formula lets you jump straight to any term — the 50th, the 100th — without listing every term in between. It also works in reverse: given a term's value, you can solve for its position n, which is exactly how "which term of the AP is…" questions are solved.

💡 Quick check: Find the 10th term of the AP 5, 8, 11, 14, … . Here a = 5, d = 3. a₁₀ = 5 + (10−1)(3) = 5 + 27 = 32.

4. Sum of First n Terms — Two Formulas

Sₙ = n/2 [2a + (n − 1)d] Use this when you know the first term, common difference, and n
Sₙ = n/2 (a + l) Use this shortcut when the last term l is already known — much faster

Both formulas give the same answer; the second is simply a shortcut once you already know (or can quickly find) the last term. The nth term itself can also be extracted from consecutive sums: aₙ = Sₙ − S₍ₙ₋₁₎, with the special case a₁ = S₁ (not S₁ − S₀).

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5. Six Fully Solved Examples

Example 1 — Finding the nth term

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

a = 3, d = 4. a₁₅ = 3 + (15−1)(4) = 3 + 56 = 59.

Example 2 — Sum of the first n terms

Find the sum of the first 20 terms of the AP 2, 7, 12, 17, …

a = 2, d = 5, n = 20. S₂₀ = 20/2 [2(2) + 19(5)] = 10[4 + 95] = 10 × 99 = 990.

Example 3 — Which term equals a given value

Which term of the AP 21, 18, 15, … is −81?

a = 21, d = −3. −81 = 21 + (n−1)(−3) → (n−1)(−3) = −102 → n−1 = 34 → n = 35 (the 35th term).

Example 4 — Working backward from Sₙ

The sum of the first n terms of an AP is Sₙ = 3n² + 5n. Find the AP.

S₁ = 3+5 = 8 = a₁. S₂ = 12+10 = 22. a₂ = S₂ − S₁ = 14. d = a₂ − a₁ = 6. The AP is 8, 14, 20, 26, …

Example 5 — Word problem (production increase)

A factory's yearly production increases by a fixed number every year. It produced 600 units in the 3rd year and 700 units in the 7th year. Find the production in the 1st year and the total production in the first 7 years.

a₃ = a + 2d = 600, a₇ = a + 6d = 700. Subtracting: 4d = 100 → d = 25 → a = 600 − 50 = 550 (1st-year production).
S₇ = 7/2[2(550) + 6(25)] = 7/2[1100+150] = 7/2 × 1250 = 4375 units in the first 7 years.

Example 6 — Three numbers in AP

The sum of three numbers in AP is 24, and their product is 440. Find the numbers.

Let the numbers be (a−d), a, (a+d). Sum: 3a = 24 → a = 8. Product: (8−d)(8)(8+d) = 440 → 8(64−d²) = 440 → 64−d² = 55 → d² = 9 → d = ±3. The numbers are 5, 8, and 11.

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6. Common Mistakes Students Make

⚠️ Mistake 1 — Off-by-one error. Writing aₙ = a + nd instead of a + (n−1)d. This single slip is the single most common error in the entire chapter — always double-check the "n−1", especially under time pressure.
⚠️ Mistake 2 — Sign errors with a negative d. In a decreasing AP (like 21, 18, 15…), forgetting that d is negative and using +3 instead of −3 flips the entire answer.
⚠️ Mistake 3 — Using aₙ = Sₙ − S₍ₙ₋₁₎ incorrectly for the first term. a₁ is simply S₁, not S₁ − S₀ (since S₀ = 0 by definition, this usually still works, but students often skip verifying it).
⚠️ Mistake 4 — Accepting a non-integer or negative value of n. The number of terms must always be a positive whole number. If solving for n gives a fraction or a negative value, re-check the setup — a valid AP question never ends with "n = 4.5".
⚠️ Mistake 5 — Wrong setup for "three/four numbers in AP" problems. Using a, a+d, a+2d instead of the symmetric a−d, a, a+d makes the algebra far messier. The symmetric form is always faster for odd-numbered sets.
⚠️ Mistake 6 — Mixing up AP and GP formulas under exam pressure. AP uses (n−1)d (addition); GP uses r^(n−1) (multiplication). Confusing the two is a common last-minute slip in combined revision.
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7. Quick Formula Recap Table

ConceptFormula
General APa, a+d, a+2d, …
Common differenced = aₙ − a₍ₙ₋₁₎
nth termaₙ = a + (n − 1)d
nth term from the end (last term l)aₙ = l − (n − 1)d
Sum of n terms (using d)Sₙ = n/2 [2a + (n − 1)d]
Sum of n terms (using last term)Sₙ = n/2 (a + l)
nth term from Sₙaₙ = Sₙ − S₍ₙ₋₁₎
Number of termsn = [(l − a) / d] + 1
Arithmetic Mean of a & bA = (a + b) / 2
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8. Shortcuts & How to Revise

Spot the last term before choosing a formula. If l is given (or easy to find), Sₙ = n/2(a+l) is almost always faster than expanding 2a+(n−1)d.
✍️ Say the nth term formula out loud as a rhythm. "a plus, n minus one, times d" — this prevents the classic n vs. (n−1) slip under time pressure.
🧮 For word problems, write the two "known" equations first. Example 5 above shows this — set up a+2d=600 and a+6d=700 before touching the sum formula.
🎯 Sanity-check your value of n. If n isn't a positive whole number, something upstream is wrong — recheck before submitting the answer.
🔁 Revisit this page before your unit test. Bookmark it and use the Quick Formula Recap Table (Section 7) as your final five-minute revision before the exam.
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9. Frequently Asked Questions

What is an Arithmetic Progression (AP)?

An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is always the same, called the common difference (d). Example: 2, 5, 8, 11, 14 is an AP with d = 3.

What is the formula for the nth term of an AP?

aₙ = a + (n − 1)d, where a is the first term, d is the common difference, and n is the term's position in the sequence.

What is the formula for the sum of n terms of an AP?

Sₙ = n/2 [2a + (n − 1)d]. If the last term l is already known, the shortcut Sₙ = n/2 (a + l) is quicker.

How do you find the common difference of an AP?

Subtract any term from the term that comes right after it: d = aₙ − a₍ₙ₋₁₎. It must give the same value for every consecutive pair in the sequence.

Can the common difference be negative or a fraction?

Yes. A negative d produces a decreasing AP (like 20, 15, 10, 5), and d can equally be a fraction or a decimal — the only rule is that it stays constant throughout.

What is the difference between an AP and a GP?

An AP has a constant difference between consecutive terms (addition-based). A Geometric Progression (GP) has a constant ratio between consecutive terms (multiplication-based). Don't mix up (n−1)d with r^(n−1) under exam pressure.

How do you find how many terms an AP has?

Rearrange the nth term formula: n = [(l − a) / d] + 1, where l is the last term. The result must always be a positive whole number.

Conclusion: Two Formulas, Endless Applications

Arithmetic Progressions is a short chapter with a long reach — the same two formulas (nth term and sum of n terms) resurface in mensuration, statistics, and even physics problems in higher classes. Once the "a + (n−1)d" rhythm and the two sum formulas are automatic, what looks like a wordy, intimidating problem usually collapses into two or three lines of substitution.

Rework the six solved examples above without checking the answers first, say the nth term formula out loud until it's second nature, and always sanity-check that your value of n makes sense. That sequence — understand, rework, verify — is what actually converts a formula sheet into exam marks.

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Tags: Arithmetic Progressions Class 10, nth Term Formula, Sum of AP, Common Difference, AP Word Problems, CBSE 2026, NCERT Class 10 Maths Chapter 5

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