A sequence is an ordered list of numbers following some rule. A series is what you get when you add the terms of that sequence. That distinction matters — many students conflate the two and then get confused when a question asks for the "sum of terms" versus "the nth term."
Think of sequences like a staircase. In an Arithmetic Progression (AP), every step is the same height — the common difference d is constant. You climb uniformly: 2, 5, 8, 11… each step adds 3. In a Geometric Progression (GP), the staircase multiplies — each step is a fixed ratio r times the previous one: 3, 6, 12, 24… each step doubles.
The real-world analogy that locks this in: AP is like a fixed monthly salary increment (add the same amount each year). GP is like compound interest (multiply by the same factor each year). NDA doesn't test this analogy directly, but it immediately tells you which formula to reach for.
For NDA specifically, AP questions dominate. Look at the PYQs — every sequence question in recent years has been AP-based. GP questions typically appear in 1-2 problems focused on sum to infinity or the relationship between AM and GM. Budget your revision time accordingly: 70% AP, 30% GP.
The deeper insight that most coaching classes skip: the nth term formula and the sum formula are not independent. If you know Sₙ, you can derive aₙ using aₙ = Sₙ − Sₙ₋₁. This relationship is a full question type by itself (see the first PYQ below). Internalise it now.
An AP with first term a and common difference d has:
aₙ = a + (n−1)dSₙ = n/2 × [2a + (n−1)d] or equivalently Sₙ = n/2 × (a + l) where l is the last termKey structural property — symmetry around the middle term:
In any AP, terms equidistant from a reference term sum to twice that reference term. Formally: if the pth term is k, then a_{p+q} + a_{p−q} = 2k. This is not a coincidence — it follows directly from the linear structure of AP. The (p+q)th and (p−q)th terms sit symmetrically on either side of the pth term, so their average equals the pth term.
This property appears in NDA questions regularly. The moment you see "sum of the pth, (p+q)th, and (p−q)th terms," don't expand — use symmetry directly.
Deriving aₙ from Sₙ:
If you're given Sₙ (sum of first n terms), then:
aₙ = Sₙ − Sₙ₋₁ for n ≥ 2a₁ = S₁This is a standard NDA question type. Don't memorise a result — derive it fresh each time using this subtraction.
Special results worth memorising:
n(n+1)/2n(n+1)(2n+1)/6[n(n+1)/2]²A GP with first term a and common ratio r has:
aₙ = a · r^(n−1)Sₙ = a(rⁿ − 1)/(r − 1) for r ≠ 1; Sₙ = na for r = 1|r| < 1): S∞ = a/(1 − r)GP symmetry: If three terms a, b, c are in GP, then b² = ac. Use this when you see three terms in GP — substitute a/r, a, ar for three terms in GP to simplify algebra dramatically.
a and b: (a + b)/2a and b: √(ab)AM ≥ GM (equality only when a = b)NDA has occasionally tested AM-GM inequality to find maximum/minimum values. Recognise the pattern: if x + y = constant, then xy is maximised when x = y.
If you insert n arithmetic means between a and b, the common difference is d = (b − a)/(n + 1).
If you insert n geometric means between a and b, the common ratio is r = (b/a)^(1/(n+1)).
Here's the pattern that shows up in NDA constantly: if Sₙ = An² + Bn, then:
a₁ = A + B (just substitute n = 1)d = 2A (the coefficient of n² doubled gives the common difference)aₙ = (2A)n + (B − A) = 2An + B − AYou can verify: aₙ = Sₙ − Sₙ₋₁ = A[n² − (n−1)²] + B[n − (n−1)] = A(2n−1) + B. So aₙ = 2An + (B − A). This is a linear function of n, confirming AP.
For Sₙ = 3n² + 5n: aₙ = 6n + 2. Set equal to 68, solve for n. This is exactly what the PYQ tests.
When a question asks for the sum of terms equidistant from a central term, use the symmetry rule instead of computing each term individually. If the pth term is k, then a_{p+q} + a_{p−q} = 2k, so the sum of all three (pth, (p+q)th, (p−q)th) is k + 2k = 3k. Standard method: expand using a + (n−1)d for three terms, collect, simplify — 6 steps, ~45 seconds. Symmetry method: recognise the pattern, write 3k — 1 step, ~5 seconds.
If Sₙ = An² + Bn + C, the nth term is aₙ = 2An + (B − A) — but only when C = 0. If C ≠ 0, the sequence is NOT a standard AP (the "series" doesn't start from a proper first term). Always check: if C = 0, use the formula directly. If C ≠ 0, compute Sₙ − Sₙ₋₁ by hand (5 steps). Forgetting to check C = 0 is one of the top error sources in this topic. Standard method with Sₙ − Sₙ₋₁ expansion: ~50 seconds. Memorised formula for C = 0 case: ~8 seconds.
If a problem says "three terms are in GP," never call them a, ar, ar². Call them a/r, a, ar. Their product is then simply a³ and their sum is a(1/r + 1 + r). This halves the algebra. Example: "three numbers in GP have product 216 and sum 19" — immediately a³ = 216 so a = 6, then solve for r. Standard substitution with a, ar, ar²: 8 algebraic steps. Substitution with a/r, a, ar: 4 steps.
When a question asks to maximise xy given x + y = S (or minimise x + y given xy = P), apply AM ≥ GM directly: (x+y)/2 ≥ √(xy), so maximum of xy = (S/2)², achieved at x = y. This appears in NDA disguised as "find the maximum product of two numbers summing to k." Standard calculus approach (differentiate, set to zero): ~60 seconds. AM-GM recognition: ~10 seconds.
When both first term a and last term l are given, use Sₙ = n/2 × (a + l) immediately — don't convert to the [2a + (n−1)d] form. You save the step of computing d. This is especially fast when n is also given explicitly. Difference: 2 steps versus 4 steps. Saves ~20 seconds on straightforward sum problems.
When you see a sequences/series question in the exam hall, run this check in order:
Step 1 — Identify the type. Is the difference between consecutive terms constant? AP. Is the ratio constant? GP. Is neither constant? Look for AGP (arithmetic-geometric progression) or special series.
Step 2 — What is being asked? nth term → use aₙ formula directly. Sum → use Sₙ formula. Given Sₙ, find aₙ → use aₙ = Sₙ − Sₙ₋₁.
Step 3 — Check for symmetry. If the question involves terms equidistant from a reference term → use the symmetry property, don't expand.
Step 4 — Three terms in GP or AP? Substitute symmetric forms (a−d, a, a+d for AP; a/r, a, ar for GP) before doing any algebra.
Step 5 — Verify with n=1 or small cases. If time allows, plug n=1 into your answer to catch arithmetic errors. Takes 5 seconds and saves wrong answers.
If the question is about sum to infinity, confirm |r| < 1 before writing S∞ = a/(1−r). Writing this formula for |r| ≥ 1 is an automatic error.
Why this question: This is the canonical "derive nth term from Sₙ" question that NDA repeats in different disguises. If you can do this mechanically in under 45 seconds, you've captured a guaranteed mark.
Solving path: Given Sₙ = 3n² + 5n. Use aₙ = Sₙ − Sₙ₋₁. Compute Sₙ₋₁ = 3(n−1)² + 5(n−1) = 3n² − 6n + 3 + 5n − 5 = 3n² − n − 2. Then aₙ = (3n² + 5n) − (3n² − n − 2) = 6n + 2. Set 6m + 2 = 68, so 6m = 66, giving m = 11. Answer: C.
Why this question: This tests the symmetry property of AP — a structural insight rather than formula-grinding. NDA uses this to separate students who understand AP from those who merely memorise formulas.
Solving path: The pth term is k. The (p+q)th and (p−q)th terms are symmetric about the pth term. By AP symmetry: a_{p+q} + a_{p−q} = 2 × a_p = 2k. Adding the pth term: total sum = k + 2k = 3k. Answer: B. You should not need to write a + (n−1)d at all.
Why this question: An algebraic verification of the same symmetry property using explicit expansion — NDA occasionally tests the same concept with different framing to check whether you understand the structure or just pattern-matched.
Solving path: Let first term = a, common difference = d. pth term = a + (p−1)d = k. Sum of the three terms: [a + (p−1)d] + [a + (p+q−1)d] + [a + (p−q−1)d]. Group: = 3a + d[(p−1) + (p+q−1) + (p−q−1)] = 3a + d[3p − 3] = 3a + 3(p−1)d = 3[a + (p−1)d] = 3k. Answer: B. Note how the q and −q cancel — this is exactly the symmetry property in algebraic dress.
Using aₙ = Sₙ/n — this gives the average of the first n terms, not the nth term. The correct relation is aₙ = Sₙ − Sₙ₋₁. This error appears in roughly one in four student solutions to Sₙ-type problems.
Forgetting to check n = 1 separately when deriving aₙ from Sₙ. The formula aₙ = Sₙ − Sₙ₋₁ requires n ≥ 2. For n = 1, compute a₁ = S₁ directly. If S₁ matches the formula for n ≥ 2, no issue — but if Sₙ has a constant term (C ≠ 0), the first term will not match the general formula.
Applying S∞ = a/(1−r) without checking |r| < 1. If |r| ≥ 1, the series diverges and has no finite sum. NDA has never penalised this in a question where |r| < 1 is clearly satisfied, but a poorly designed distractor can tempt you to apply it blindly.
Using a, ar, ar² for three terms in GP — then struggling with the product being a³r³ rather than a clean cube. Always substitute a/r, a, ar when three GP terms are given. The product is immediately a³.
Confusing common difference with the nth term formula. If Sₙ = An², students sometimes say d = A. The correct result is d = 2A. Derive it: aₙ = Sₙ − Sₙ₋₁ = A[n² − (n−1)²] = A(2n−1), so d = aₙ − aₙ₋₁ = 2A.
Summing "n terms" vs "first n terms." When a question says "the 5th to 12th terms," compute S₁₂ − S₄, not S₁₂ − S₅. The first 4 terms are excluded, not the first 5.