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Sequences and Series hero

Sequences and Series

~14 min read

In 30 seconds
  • What: Sequences and Series covers Arithmetic Progression (AP), Geometric Progression (GP), Harmonic Progression (HP), their means, and special sums — all tested heavily in NDA Mathematics Paper.
  • Why it matters: This topic has appeared consistently from 2010 to 2023, with 4–8 questions per paper. Mastering it is a direct path to guaranteed marks.
  • Key fact: \(\text{AM} \ge \text{GM} \ge \text{HM}\) for any two positive numbers, and the relationship \(\text{AM} \times \text{HM} = \text{GM}^2\) is a recurring NDA formula.

Sequences and Series is one of the most predictable and high-scoring topics in NDA Mathematics. The NDA paper has asked questions on this topic every single year from 2010 to 2023. Unlike some chapters where difficulty spikes unpredictably, this topic follows a clear pattern: identify the type of progression, apply the right formula, and reach the answer in under two minutes. Build your toolkit here and bank 4–8 marks per paper with confidence.

What This Topic Covers

This topic is broader than just memorising formulas for AP and GP. NDA tests your ability to switch between progressions, recognise disguised sequences, and use means strategically. Here is a clear map of everything under this chapter.

Sub-topics tested in NDA

  • Arithmetic Progression (AP) — nth term, sum of n terms, inserting means
  • Geometric Progression (GP) — nth term, sum of finite and infinite GP, product of terms
  • Harmonic Progression (HP) — nth term, harmonic mean, conditions for HP
  • Arithmetic Mean (AM), Geometric Mean (GM), Harmonic Mean (HM) — individual and combined
  • AM–GM–HM inequality and the relationship \(\text{AM} \times \text{HM} = \text{GM}^2\)
  • Arithmetico-Geometric Progression (AGP) — sum of AGP series
  • Special sums — sum of natural numbers, sum of squares, sum of cubes
  • Logarithms in progressions — recognising when log terms form an AP or GP
  • Common terms between two APs and two GPs

The NDA paper does not restrict questions to a single sub-type. A question from 2016 (NDA 2016-II) combined AP and HP in a single problem block, asking for the value of ab when a, x, y, z, b are in AP and a, p, q, r, b are in HP with given sum conditions. Expect cross-type questions regularly.

Exam Pattern & Weightage

The table below is built directly from the PYQ file. It shows when questions appeared and the dominant sub-type tested in each year. Use this to prioritise your preparation — the topics that appear most frequently deserve the most attention.

Year Paper No. Main Sub-types Tested
2010 I & II 3 AM–GM–HM relationship, infinite GP
2011 I & II 10 AP, GP, HM, common terms, GP ratio, log in AP
2012 I & II 9 AP sum, GM of sequence, HP, special sums, infinite GP
2013 I & II 5 AP n-th term, GP condition, infinite GP, AP sum property
2014 I & II 5 AP with function, S3n/Sn ratio, sum of AP blocks
2015 I & II 5 Repeating decimals, infinite products, GP means
2016 I & II 9 AP+HP combined, GM of logs, polygon angles in AP
2017 I & II 10 AM/GM ratio, infinite GP, sum of odd two-digit numbers, common difference
2018 I & II 9 AM:GM ratio, HP condition, GP product, infinite GP sum
2019 I & II 8 GP n-th term, AP insertion, sum of (p+q)th term, HM condition
2020 I 2 GP power condition, logs in AP
2021 I & II 8 S30/S20-S10, GP multiplication property, HM condition
2022 I & II 6 Sum of AP with ratio, GP properties, HP condition, sum of even numbers
2023 I & II 9 Log in GP, AP largest negative term, GM of powers, HM of sequence
NDA Alert

NDA 2011 alone had 10 questions from this chapter across both papers. Any year can be high-volume. Never treat this topic as "partially prepared."

Core Concepts

Work through each sub-type in order. Each formula card shows the exact result you need to recall. Do not try to re-derive these in the exam — just apply them.

Arithmetic Progression (AP)

An AP is a sequence where the difference between consecutive terms is constant. That constant is the common difference d.

AP — nth Term $$T_n = a + (n-1)d$$ where \(a\) = first term, \(d\) = common difference
AP — Sum of n Terms $$S_n = \tfrac{n}{2}\bigl[2a + (n-1)d\bigr] = \tfrac{n}{2}(\text{first term} + \text{last term})$$
AP — Common Difference from Sum Formula If $$S_n = nP + \tfrac{n(n-1)Q}{2}$$ then \(d = Q\) (NDA 2017-II tested this directly)

Key property tested in NDA 2011-II (Q.15): If \(a, b, c, d, e, f\) are in AP, then \((e - c) = 2(d - c) = 2d\) (where \(d\) is the common difference). The NDA often tests such "difference of terms" identities without explicitly naming them.

NDA Alert

When \(S_5 = S_{10}\) for an AP (NDA 2013-II), the common difference must be negative. This is a statement-type question — the reasoning is that \(S_{10} - S_5 = 0\) means terms 6 through 10 sum to zero, which forces \(d\) to be negative when \(a > 0\) is not guaranteed. The correct statement is that either \(a\) or \(d\) is negative, but not necessarily both.

Geometric Progression (GP)

A GP has a constant ratio r between consecutive terms. Questions on GP in NDA span from basic nth term to infinite sums, product identities, and GP conditions for logs.

GP — nth Term $$T_n = ar^{n-1}$$ where \(a\) = first term, \(r\) = common ratio
GP — Sum of n Terms (r ≠ 1) $$S_n = \dfrac{a(r^n - 1)}{r - 1}\ \text{for}\ r > 1;\quad S_n = \dfrac{a(1 - r^n)}{1 - r}\ \text{for}\ r < 1$$
GP — Sum of Infinite GP (|r| < 1) $$S_\infty = \dfrac{a}{1 - r}$$
GP — Product of n Terms Product of first \(n\) terms $$= a^n \cdot r^{n(n-1)/2}$$ Special case: if 3rd term \(= 3\), product of first 5 terms \(= 3^5 = 243\) (NDA 2018-I, 2021-I)
NDA Alert

Do not confuse "common difference" (AP) with "common ratio" (GP). If a question says "successive terms differ by a constant factor", it is a GP (ratio), not an AP (difference). When in doubt, compute $$t_2 - t_1$$ and $$t_2 / t_1$$ for the given numbers — whichever stays constant across the sequence decides the type.

NDA consistently tests the ratio of GP sums. From NDA 2014-II: if \(3S_3 = S_{6n}\), then \(S_{3n} : S_n = 8 : 1\) and \(S_{5n} : S_{2n} = 3 : 1\). Practise deriving such ratios — they appear as two-item sets.

GP — Common Ratio for In-Term Condition If any term \(= \tfrac{1}{3} \times\) (sum of next two terms), then $$ar^n = \tfrac{1}{3}(ar^{n+1} + ar^{n+2}) \implies r = \dfrac{\sqrt{3} - 1}{2}$$ (NDA 2011-I)

Harmonic Progression (HP)

An HP is a sequence whose reciprocals form an AP. NDA tests HP identification, finding HM, and conditions linking AP, GP, and HP.

HP — nth Term $$T_n = \dfrac{1}{a + (n-1)d}$$ where \(\tfrac{1}{a}, \tfrac{1}{T_2}, \dots\) form an AP
Harmonic Mean of two numbers a and b $$\text{HM} = \dfrac{2ab}{a + b}$$

From NDA 2012-I (Q.19): if \(\tfrac{1}{4}, \tfrac{1}{x}, \tfrac{1}{10}\) are in HP, then \(4, x, 10\) are in AP, so \(x = 7\). This is the standard HP identification trick — flip all terms, check AP.

Means and the AM–GM–HM Inequality

For any two positive numbers \(a\) and \(b\):

AM, GM, HM Formulas $$\text{AM} = \dfrac{a+b}{2}\ \ \Big|\ \ \text{GM} = \sqrt{ab}\ \ \Big|\ \ \text{HM} = \dfrac{2ab}{a+b}$$
Key Inequality $$\text{AM} \ge \text{GM} \ge \text{HM}$$ with equality when \(a = b\)
AM–GM–HM Product Rule $$\text{GM}^2 = \text{AM} \times \text{HM}$$ (i.e., GM is the geometric mean of AM and HM)
Quadratic with a, b as Roots (using A and G) $$x^2 - 2Ax + G^2 = 0$$ where $$A=\tfrac{a+b}{2}$$ and $$G=\sqrt{ab}$$
NDA Alert

AM = GM (equivalently GM = HM) holds only when the two numbers are equal. If a question states $$A=G$$ for two positive numbers, the immediate conclusion is $$a=b$$ — no further algebra needed. Conversely, if $$a \neq b$$, the inequality is strict: $$A > G > H$$, never equal.

NDA 2010-I tested this directly: AM exceeds GM by 2, GM exceeds HM by 1.6. Using \(\text{AM} \times \text{HM} = \text{GM}^2\) along with the given differences leads to the two numbers being 16 and 4. This combined-means approach is a favourite NDA pattern.

From NDA 2012-I (Q.21): \(\text{GM} = 10\), \(\text{HM} = 8\) for two non-negative numbers. Then \(\text{AM} = \tfrac{\text{GM}^2}{\text{HM}} = \tfrac{100}{8} = 12.5\). The answer is 12.5 — and the NDA choices confirmed this. Always use the product rule first when both GM and HM are given.

From NDA 2018-I (Q.78): \(\text{AM} : \text{GM} = 5 : 3\). Let \(\text{AM} = 5k\), \(\text{GM} = 3k\). Then \(\text{AM}^2 - \text{GM}^2 = \tfrac{(a-b)^2}{4}\), giving \(a : b = 9 : 1\).

Arithmetico-Geometric Progression (AGP)

An AGP has terms that are products of corresponding terms of an AP and a GP. NDA occasionally uses AGP language (NDA 2014-II, Q.43-44) to describe derivative-based sequences. The sum of an AGP series is computed by the standard formula but is rarely tested directly in NDA — recognising AGP when the options list it as a distractor matters more.

Special Series Sums

Sum of First n Natural Numbers $$1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}$$
Sum of Squares of First n Natural Numbers $$1^2 + 2^2 + \dots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$$ For \(n = 20\): 2870 (NDA 2012-I confirmed this)
Sum of Cubes of First n Natural Numbers $$1^3 + 2^3 + \dots + n^3 = \left[\dfrac{n(n+1)}{2}\right]^2$$ For \(n = 20\): $$\left[\tfrac{20 \times 21}{2}\right]^2 = 210^2 = 44100.$$ NDA 2012-I noted the sum of cubes of first 20 natural numbers is 44100, not 44400 — statement 1 in Q.25 was false.
Sum of All Two-Digit Odd Numbers First \(= 11\), last \(= 99\), \(n = 45\) terms. $$S = \tfrac{45}{2}(11 + 99) = 2475$$ (NDA 2017-I)

Logs and Progressions

When log terms appear in a progression question, convert the log condition into an algebraic one using basic log properties. From NDA 2011-II (Q.13) and NDA 2021-II (Q.113): if \(\log_{10} 2,\ \log_{10}(2^x - 1),\ \log_{10}(2^x + 3)\) are in AP, then \(2(2^x - 1) = 2 + (2^x + 3)\), which gives \(x = \log_2 5\).

Log in GP Condition If \(\log x, \log y, \log z\) are in AP, then \(x, y, z\) are in GP (NDA 2016-I, 2021-II)

Worked Examples

Each example below is taken from the PYQ file. The solution path matches what the NDA expects in under two minutes.

Example 1 — AM, GM, HM Combined (NDA 2010-I)

Question: The AM of two numbers exceeds their GM by 2 and the GM exceeds their HM by 1.6. Find the two numbers.

  • Let \(\text{AM} = m\), \(\text{GM} = g\), \(\text{HM} = h\). Given: \(m - g = 2\) and \(g - h = 1.6\).
  • Use \(\text{AM} \times \text{HM} = \text{GM}^2\): \(m \cdot h = g^2\).
  • Let \(g = k\). Then \(m = k + 2\) and \(h = k - 1.6\).
  • $$(k+2)(k-1.6) = k^2 \implies k^2 + 0.4k - 3.2 = k^2 \implies 0.4k = 3.2 \implies k = 8.$$
  • \(\text{GM} = 8\), \(\text{AM} = 10\). So \(\tfrac{a+b}{2} = 10 \implies a + b = 20\), and \(\sqrt{ab} = 8 \implies ab = 64\).
  • Solving \(a + b = 20\), \(ab = 64\): \((a - b)^2 = 400 - 256 = 144\), so \(a - b = 12\). The numbers are 16 and 4. Answer: (a).

Example 2 — Common Terms Between Two APs (NDA 2011-II)

Question: What is the 10th common term between the series 2 + 6 + 10 + ... and 1 + 6 + 11 + ...?

  • Series 1: \(a = 2, d = 4\). Series 2: \(a = 1, d = 5\).
  • Common terms also form an AP. First common term: 6 (appears in both).
  • Common difference of the new AP \(= \text{LCM}(4, 5) = 20\).
  • Common AP: \(6, 26, 46, \dots\), $$T_n = 6 + (n - 1) \times 20.$$
  • $$T_{10} = 6 + 9 \times 20 = 6 + 180 = 186.$$ Answer: (b).

Example 3 — Sum of n Terms and Common Difference (NDA 2012-II)

Question: The sum of the first 10 terms and 20 terms of an AP are 120 and 440 respectively. Find the first term and common difference.

  • $$S_{10} = \tfrac{10}{2}(2a + 9d) = 5(2a + 9d) = 120 \implies 2a + 9d = 24 \quad \text{(i)}$$
  • $$S_{20} = \tfrac{20}{2}(2a + 19d) = 10(2a + 19d) = 440 \implies 2a + 19d = 44 \quad \text{(ii)}$$
  • Subtract (i) from (ii): \(10d = 20 \implies d = 2\).
  • Substitute into (i): \(2a + 18 = 24 \implies 2a = 6 \implies a = 3\).
  • First term \(= 3\) (answer b for Q.27), common difference \(= 2\) (answer b for Q.28).

Example 4 — GP: 10th Term and 4th Term Given (NDA 2011-II)

Question: If the 10th term of a GP is 9 and the 4th term is 4, what is its 7th term?

  • \(T_{10} = ar^9 = 9\) and \(T_4 = ar^3 = 4\).
  • Divide: \(r^6 = \tfrac{9}{4}\), so \(r^3 = \tfrac{3}{2}\).
  • $$T_7 = ar^6 = (ar^3) \cdot r^3 = 4 \cdot \tfrac{3}{2} = 6.$$
  • Answer: (a) 6. Note how the 7th term is the geometric mean of the 4th and 10th terms — a standard GP property NDA tests repeatedly.

Example 5 — HP via "Flip to AP" Method

Question: If the 3rd term of an HP is $$\tfrac{1}{7}$$ and the 7th term is $$\tfrac{1}{15}$$, find the 10th term.

  • Flip the terms to convert HP to AP: the 3rd term of the AP is 7, and the 7th term of the AP is 15.
  • Set up: $$a + 2d = 7$$ and $$a + 6d = 15$$.
  • Subtract: $$4d = 8 \implies d = 2$$. Then $$a + 4 = 7 \implies a = 3$$.
  • 10th term of the AP: $$a + 9d = 3 + 18 = 21$$.
  • Flip back: the 10th term of the HP is $$\tfrac{1}{21}$$. This "flip → solve as AP → flip back" routine is the standard HP method NDA expects.

Example 6 — Bouncing Ball as Two Infinite GPs

Question: A ball is dropped from a height of 100 metres. After each bounce, it returns to 80% of its previous height. Find the total distance travelled before coming to rest.

  • Downward distances form an infinite GP: $$100, 80, 64, \dots$$ with $$a=100, r=0.8$$.
  • Upward distances form an infinite GP: $$80, 64, 51.2, \dots$$ with $$a=80, r=0.8$$.
  • Total downward sum: $$S_\infty = \dfrac{100}{1-0.8} = 500$$ m.
  • Total upward sum: $$S_\infty = \dfrac{80}{1-0.8} = 400$$ m.
  • Total distance = $$500 + 400 = 900$$ m. Whenever a "bouncing ball" or "swinging pendulum" word problem appears, split into two GPs (down + up) and apply $$S_\infty$$ to each.

Example 7 — Infinite GP Sum to Identify First Term (NDA 2010-II)

Question: The sum of an infinite GP is 6. The sum of the first two terms is 9/2. What is the first term?

  • $$S_\infty = \dfrac{a}{1 - r} = 6 \implies a = 6(1 - r) \quad \text{(i)}$$
  • Sum of first two terms: $$a + ar = a(1 + r) = \tfrac{9}{2} \quad \text{(ii)}$$
  • Substitute (i) into (ii): $$6(1 - r)(1 + r) = \tfrac{9}{2} \implies 6(1 - r^2) = \tfrac{9}{2} \implies 1 - r^2 = \tfrac{3}{4} \implies r^2 = \tfrac{1}{4} \implies r = \tfrac{1}{2}$$ (taking positive root for convergence).
  • From (i): \(a = 6(1 - \tfrac{1}{2}) = 3\). Answer: (d) 3 (or 9 if \(r = -\tfrac{1}{2}\) is also valid, giving \(a = 9\)).

Exam Shortcuts (Pro-Tips)

Sequences and Series rewards pattern recognition. The five shortcuts below collapse multi-step algebra into a single move and have each appeared in at least one PYQ. Drill them until they feel automatic.

Shortcut 1 — Assume Simple Values for "If a, b, c are in AP/GP/HP"

When a question begins with "If a, b, c are in A.P./G.P./H.P." and asks for a complex algebraic relation, do not solve symbolically. Plug in the smallest valid triple and test the options.

Plug-in Triples AP: $$a=1, b=2, c=3$$  ·  GP: $$a=1, b=2, c=4$$  ·  HP: $$a=2, b=3, c=6$$ (so $$\tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{6}$$ are in AP)

Shortcut 2 — Symmetric Terms Trick

When the sum or product of three or four consecutive terms is given, choose terms symmetrically so the variable $$a$$ cancels in the sum (AP) or product (GP).

Symmetric Choices 3 in AP: $$a-d,\ a,\ a+d$$  ·  4 in AP: $$a-3d,\ a-d,\ a+d,\ a+3d$$  ·  3 in GP: $$\tfrac{a}{r},\ a,\ ar$$

Shortcut 3 — Ratio of Sums to Ratio of Terms

If the ratio of sums of $$n$$ terms of two APs is given as a function of $$n$$, the ratio of their $$m$$-th terms is obtained by replacing $$n$$ with $$2m-1$$.

Sum-Ratio to Term-Ratio If $$\dfrac{S_n}{S'_n} = \dfrac{7n+1}{4n+27}$$, then $$\dfrac{T_{11}}{T'_{11}} = \dfrac{7(21)+1}{4(21)+27} = \dfrac{148}{111} = \dfrac{4}{3}$$

Shortcut 4 — AM ≥ GM ≥ HM with Equality Test

The chain $$\text{AM} \geq \text{GM} \geq \text{HM}$$ holds for any two positive numbers, with equality iff $$a=b$$. Whenever a question gives any two of the means, the third follows from $$G^2 = A \cdot H$$ — never set up two-variable equations from scratch.

Product Rule (NDA favourite) $$G^2 = A \times H \implies A = \dfrac{G^2}{H},\quad H = \dfrac{G^2}{A}$$

Shortcut 5 — Summation Identities

Any sum involving consecutive integers, squares, or cubes reduces to one of three closed forms. NDA hides these inside word problems (sum of two-digit odd numbers, sum of cubes for $$n=20$$, etc.).

Three Must-Recall Sums $$\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}\ \ |\ \ \sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}\ \ |\ \ \sum_{k=1}^{n} k^3 = \left[\dfrac{n(n+1)}{2}\right]^2$$

Shortcut 6 — Infinite GP Convergence Check

The formula $$S_\infty = \dfrac{a}{1-r}$$ is valid only when $$|r| < 1$$. If $$|r| \geq 1$$, the series diverges and the question has no finite answer — but NDA still lists numeric distractors. Always check $$|r|$$ first.

NDA Alert

If a problem gives $$S_\infty = 6$$ and $$a + ar = \tfrac{9}{2}$$ and you solve and get $$r = \pm \tfrac{1}{2}$$, both roots are valid (both satisfy $$|r|<1$$). The two corresponding first terms — $$a=3$$ for $$r=\tfrac{1}{2}$$ and $$a=9$$ for $$r=-\tfrac{1}{2}$$ — may both appear in the answer choices. Read the option list before discarding either root.

Common Question Patterns

NDA uses the same structural patterns across years. Once you recognise the pattern, you can solve even unfamiliar-looking questions quickly.

Pattern 1 — "Which progression?" identification

Given a condition on \(a, b, c\) (or involving a quadratic equation), identify whether they are in AP, GP, or HP. Example: NDA 2017-I asked if the sum of roots equals the sum of reciprocals of squares, are \(a, ca^2, c^2\) in AP or GP? The answer was GP. Method: convert the verbal condition into algebra, then check ratios or differences.

Pattern 2 — Two-item sets on AP sum ratios

NDA frequently groups two questions around a common premise. NDA 2014-II grouped three questions on \(f(x) = ax^2 + bx + c\) where \(a, b, c\) are in AP. NDA 2016-II grouped questions on AP and HP with the same endpoints. Recognise the common premise and extract what you need for each sub-question separately.

Pattern 3 — Repeating decimals as infinite GP

\(0.9 + 0.09 + 0.009 + \dots\) is an infinite GP with first term \(0.9\) and \(r = 0.1\). Sum \(= \tfrac{0.9}{1 - 0.1} = 1\). NDA 2013-II asked this directly. Similarly, \(0.3 + 0.33 + 0.333 + \dots\) can be written as \(\tfrac{1}{3}(0.9 + 0.99 + \dots)\) and solved via the formula for sum of \(n\) terms.

Pattern 4 — GM of a GP sequence

For a sequence \(2, 4, 8, 16, 32\) (which is a GP with \(a = 2, r = 2\)), the \(\text{GM} = (\text{product})^{1/5} = (2 \times 4 \times 8 \times 16 \times 32)^{1/5} = (2^{15})^{1/5} = 2^3 = 8\) (NDA 2011-I). The shortcut: GM of a GP \(=\) middle term when \(n\) is odd.

NDA Alert

Product of first 5 terms of a GP when the 3rd term is 3: answer is always \(3^5 = 243\). This appeared in both NDA 2018-I and NDA 2021-I. You do not need the first term or common ratio — the middle term raised to the power \(n\) gives the product of \(n\) terms in a GP.

Pattern 5 — Polygon angles in AP

Interior angles of an \(n\)-sided polygon in AP appeared in NDA 2012-II (angles in AP, least \(= 30°\), greatest \(=\) ?) and NDA 2016-II (least \(= 120°\), \(d = 5°\)). The constraint is: sum of interior angles \(= (n - 2) \times 180°\). With the AP sum formula, set up and solve for \(n\). Watch out: \(n\) must be a positive integer, so not all solutions are valid.

How NDA Tests This Topic

About 40% of questions are direct formula application (find nth term, find sum). About 35% require recognising which progression two or three numbers satisfy. The remaining 25% are multi-step: combining AP and HP, working with logs in progressions, or using the AM–GM–HM chain. Build speed on the direct formula questions so you have time for multi-step ones.

Preparation Strategy

This topic rewards systematic preparation more than any other in NDA Maths. Here is a week-by-week approach.

Day 1–2: Formulas and basic application

Write every formula card in this article on index cards. Drill the nth term and sum formula for AP and GP until you can apply them without looking. Test yourself on: finding a term given two others, finding sum given first and last term, and computing AM, GM, HM for given numbers.

Day 3–4: HP, means, and combined problems

Practise the "flip to AP" trick for HP until it is automatic. Memorise AM × HM = GM². Work through NDA 2010-I, 2012-I, 2013-I and 2018-I questions that combine two or three means in a single problem.

Day 5–6: Special series and disguised progressions

Practise summing 1² + 2² + ... + n² and 1³ + 2³ + ... + n³. Practise recognising when a log condition implies a GP or AP. Work through the repeating decimal problems (0.9 + 0.09 +... style). Do the polygon-in-AP questions from 2012-II and 2016-II.

Day 7: Full PYQ drill

Solve a timed set of 20 questions from the PYQ file under exam conditions (25 seconds per question target). Mark anything that took over 90 seconds and revisit the formula or method involved. The goal is to reach a point where pattern recognition, not calculation, drives most answers.

Must-remember shortcuts

  • Middle term of a GP \(=\) GM of all terms when \(n\) is odd
  • If \(a, b, c\) are in GP: \(b^2 = ac\)
  • If \(a, b, c\) are in HP: \(b = \tfrac{2ac}{a + c}\)
  • If \(a, b, c\) are in AP: \(2b = a + c\)
  • \(S_n\) (sum formula) gives the nth term as \(T_n = S_n - S_{n-1}\) for \(n \ge 2\)
  • Common terms of two APs form a new AP with common difference \(=\) LCM of the two CDs
  • If \(p, q, r\) are in both AP and GP, then \(p = q = r\) (NDA 2012-I, Q.20)

Link this topic to Binomial Theorem and Logarithms — log-based progression questions will make more sense once you are fluent in log rules. Also connect to Quadratic Equations — NDA often asks when roots of a quadratic are in AP, GP, or HP, and you need both skill sets simultaneously.

Test Your Sequences and Series Skills

Defence Road mock tests include topic-wise drills on AP, GP, HP and special series — with instant solutions that show the fastest method, not just the answer.

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Frequently Asked Questions

How many questions from Sequences and Series appear in NDA Maths?

Based on the PYQ record from 2010 to 2023, this topic contributes between 4 and 10 questions per paper. Years like 2011 and 2017 had 10 questions. On average, expect 5–7 questions. That makes it one of the top three contributors to your NDA Maths score.

What is the most important formula in this chapter for NDA?

The \(\text{AM} \times \text{HM} = \text{GM}^2\) relationship is the single most reusable formula across NDA questions involving means. It lets you find the third mean when two are given, and it powers the solution for combined-means problems that appear almost every year. After that, the infinite GP sum formula \(S_\infty = \tfrac{a}{1 - r}\) is tested in multiple disguised forms (repeating decimals, infinite products, alternating series).

How do I identify whether a, b, c are in AP, GP, or HP from a word problem?

Check three conditions: if \(2b = a + c\) \(\to\) AP; if \(b^2 = ac\) \(\to\) GP; if \(\tfrac{2}{b} = \tfrac{1}{a} + \tfrac{1}{c}\) (equivalently \(2ac = b(a + c)\)) \(\to\) HP. When a condition is given in terms of roots of equations or sums/products, convert it algebraically to one of these three forms. NDA 2017-I and 2018-II both used this approach.

What is the trick for "product of first n terms of a GP" problems?

If the 3rd term of a GP is \(k\), the product of the first 5 terms is \(k^5 = 3^5 = 243\) when \(k = 3\). This works because in a GP, the product of terms equidistant from the middle equals the square of the middle term. So the product of \(n\) terms (where \(n\) is odd) \(= (\text{middle term})^n\). This appeared in NDA 2018-I and 2021-I.

How do I handle questions where log terms are said to be in AP or GP?

If \(\log_{10} a, \log_{10} b, \log_{10} c\) are in AP, it means \(\log_{10} b - \log_{10} a = \log_{10} c - \log_{10} b\), which gives \(\log_{10}(b/a) = \log_{10}(c/b)\), so \(b/a = c/b\), meaning \(a, b, c\) are in GP. Similarly, if \(\log a, \log b, \log c\) are in GP, then \((\log b)^2 = (\log a)(\log c)\). Convert the log statement to an algebraic condition first, then work with AP, GP, or HP rules as usual.

What is the sum of the first n natural number cubes and squares — do I need to memorise them?

Yes. Sum of squares \(= \tfrac{n(n+1)(2n+1)}{6}\). Sum of cubes \(= \left[\tfrac{n(n+1)}{2}\right]^2\). NDA 2012-I (Q.25) tested both for \(n = 20\): squares give 2870 (correct) and cubes give 44100 — the paper listed 44400 as false. You need these formulas correctly, not approximately.

Can I skip HP and focus only on AP and GP?

No. HP questions appear in NDA almost every year, often combined with AM and GM in a single question block. Skipping HP means losing 1–2 questions per paper. The HP formula is just one extra thing to remember — practise the "flip to AP" method and you will find HP questions become almost as fast as AP questions.