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Some questions hand you the sum instead of the sequence: "the sum of the first n terms is n(4n + 1); find the 10th term." The method is one subtraction, but there is a trap hidden at n = 1 that catches many students. This post shows why the subtraction works, applies it to arithmetic and geometric cases, and ends with a ratio question that looks hard until you use the same idea.
Where is this in the syllabus?
NCERT's Class 10 chapter Arithmetic Progressions (2026-27 reprint) states it as a remark: "The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., an = Sn – Sn – 1." The Class 11 chapter Sequences and Series uses Sₙ for sums throughout; its current sections cover sequences, series, geometric progressions and the A.M.–G.M. relation. The JEE (Main) 2026 syllabus lists arithmetic and geometric progressions and means in Unit 6.
Why does the subtraction work?
Write the two sums out:
Sₙ = a₁ + a₂ + … + aₙ₋₁ + aₙ
Sₙ₋₁ = a₁ + a₂ + … + aₙ₋₁
Everything cancels except aₙ. Nothing in this uses a common difference, so the rule holds for any sequence. But Sₙ₋₁ only makes sense when n − 1 ≥ 1, so the formula is guaranteed only for n ≥ 2. The first term comes from a₁ = S₁.
Worked example 1: an AP hiding in a quadratic
Sₙ = n(4n + 1) = 4n² + n.
Sₙ₋₁ = 4(n − 1)² + (n − 1) = 4n² − 8n + 4 + n − 1 = 4n² − 7n + 3.
aₙ = (4n² + n) − (4n² − 7n + 3) = 8n − 3.
Check n = 1: S₁ = 5 and 8(1) − 3 = 5. They agree, so the formula holds from the first term.
The sequence is 5, 13, 21, 29, …, an AP with d = 8. The 10th term is 8(10) − 3 = 77, and S₁₀ − S₉ = 410 − 333 = 77 agrees.
A shortcut follows: whenever Sₙ = pn² + qn, the sequence is an AP with d = 2p and a₁ = p + q. Here p = 4 and q = 1, so d = 8 and a₁ = 5.
Worked example 2: the constant-term trap
Sₙ = n² + n + 1.
Sₙ₋₁ = (n − 1)² + (n − 1) + 1 = n² − n + 1.
aₙ = 2n for n ≥ 2.
But S₁ = 3, while 2(1) = 2.
The sequence is 3, 4, 6, 8, 10, …, with differences 1, 2, 2, 2. It is not an AP, although every term after the first fits one. A constant term in Sₙ is the warning sign: the subtraction cancels it, so it shows up only in a₁.
Worked example 3: geometric sums
Sₙ = 2ⁿ − 1 gives aₙ = 2ⁿ − 2ⁿ⁻¹ = 2ⁿ⁻¹, and S₁ = 1 = 2⁰. This is the GP 1, 2, 4, 8, … with r = 2.
Now try Sₙ = 4ⁿ + 2. For n ≥ 2, aₙ = 4ⁿ − 4ⁿ⁻¹ = 3 · 4ⁿ⁻¹, giving 12, 48, 192, … But S₁ = 6, not 3. The sequence 6, 12, 48, 192, … is not a GP. Sums of the form k(rⁿ − 1) give a true GP from the first term; anything else needs the n = 1 check.
Worked example 4: a ratio question
For an AP, Sₘ ÷ Sₙ = m² ÷ n² for all m and n. Find aₘ ÷ aₙ.
Sₙ ÷ n² is the same for every n, so Sₙ = kn² for some constant k.
aₙ = kn² − k(n − 1)² = k(2n − 1). Check: S₁ = k = k(2 − 1).
So aₘ ÷ aₙ = (2m − 1) ÷ (2n − 1).
For example, a₆ ÷ a₂ = 11 ÷ 3.
Which slips cost marks?
Expanding (n − 1)² as n² − 1. It is n² − 2n + 1.
Skipping the n = 1 check, as in examples 2 and 3.
Concluding that a sequence is an AP because aₙ came out linear for n ≥ 2.
Finding a₁₀ as S₁₀ alone, which is the sum, not the term.
When your subtraction keeps going wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can find the line where your algebra slipped.
A practice routine
Find aₙ and check n = 1 for Sₙ = 3n² − 2n, Sₙ = n³, Sₙ = 5ⁿ − 1 and Sₙ = 2n² + n + 4.
For each, say whether the sequence is an AP, a GP or neither, from the first term.
Do NCERT Class 10 Exercise 5.3, question 11, using the subtraction.
Redo worked example 4 for an AP with Sₙ = kn(3n + 1), and find a₅ ÷ a₂.
For the formulas behind APs, see our Class 10 guide to arithmetic progressions; for GPs and means, read sequences and series for JEE. Tonight, do step 1 and write the n = 1 check in a box every time.
