📘 Algebra · Chapter MN04🎯 NDA Level : High Priority
Sequences and Series is one of the highest-scoring chapters in NDA Mathematics. Questions are formula-driven and straightforward — making this a guaranteed scoring area with good revision. Every sub-topic (AP, GP, HP, AGP, Summation) has been tested in recent NDA papers, with AP and GP appearing almost every year.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding the nth term or sum of n terms of an AP or GP;
(2) Given two or three terms of an AP/GP, finding a, d, or r;
(3) Sum of infinite GP with |r| < 1;
(4) Inserting arithmetic or geometric means between two numbers;
(5) The AM ≥ GM inequality applied to find minimum/maximum values;
(6) HP — finding nth term using 1/HP = AP relationship;
(7) Summation of series using Σn, Σn², Σn³ standard formulae;
(8) AGP — finding sum by the standard method.
Topics at a Glance
① Arithmetic Progression
nth term, Sₙ, AM, inserting means
② Geometric Progression
nth term, Sₙ, S∞, GM, |r|<1
③ Harmonic Progression
HM, 1/HP is AP, HM formula
④ AM, GM, HM Relations
AM ≥ GM ≥ HM, AM·HM = GM²
⑤ AGP
Arithmetic-Geometric series, sum method
⑥ Standard Summation
Σn, Σn², Σn³ formulae
1. Arithmetic Progression (AP)
1.1
Definition, nth Term & Sum of n Terms
Foundation of sequences — most tested type in NDA
An Arithmetic Progression (AP) is a sequence in which each term differs from the previous by a fixed constant called the common difference (d). General form: a, a+d, a+2d, a+3d, …
⚡ Core AP Formulae
nth term: Tₙ = a + (n−1)d
Last term: l = a + (n−1)d (when l is the last of n terms)
Sum of n terms:
Sₙ = n/2 · [2a + (n−1)d]
Sₙ = n/2 · (a + l) ← use this when last term l is known
nth term from sum:
Tₙ = Sₙ − Sₙ₋₁ ← valid for n ≥ 2; T₁ = S₁
Arithmetic Mean of n numbers:
AM = (Sum of all terms) / n = (a + l) / 2 for an AP
a = first term, d = common difference, n = number of terms, l = last term. Always identify which quantities are given before choosing the right formula.
Key Properties of AP
d = Tₙ₊₁ − Tₙ (constant throughout)
If d > 0 → increasing AP
If d < 0 → decreasing AP
If d = 0 → constant sequence
Three consecutive AP terms: a−d, a, a+d (use this for problems)
Four consecutive AP terms: a−3d, a−d, a+d, a+3d
Inserting Arithmetic Means
To insert n AMs between A and B:
Common difference: d = (B−A)/(n+1)
First AM = A + d, second = A + 2d, … nth = A + nd
Single AM between A and B = (A+B)/2
Sum of n AMs = n × AM = n(A+B)/2
Worked Example — Finding n and Sum
The sum of first n terms of AP 3, 7, 11, … is 820. Find n.
a = 3, d = 4. Sₙ = n/2[2(3)+(n−1)4] = n/2[6+4n−4] = n/2[4n+2] = n(2n+1).
📌 NDA Shortcut — Three Terms in AP:
Whenever a problem says "three numbers are in AP", assume them as a−d, a, a+d. Their sum = 3a (the d cancels), making the arithmetic much simpler. Similarly, four terms in AP: assume a−3d, a−d, a+d, a+3d — sum = 4a.
1.2
Visual: AP on the Number Line
Equally spaced points — d is the constant gap
Fig 1: Arithmetic Progression — equal spacing d between each consecutive term.
📋 TOPIC-WISE PYQ
Arithmetic Progression — NDA-Pattern Questions
Q1. The 10th term of the AP 2, 5, 8, 11, … is:
(a) 28 (b) 29 (c) 30 (d) 32
Answer: (b) 29
a = 2, d = 3, n = 10.
T₁₀ = 2 + (10−1)×3 = 2 + 27 = 29.
Q2. How many terms of the AP 3, 5, 7, … must be taken so that the sum is 120?
(a) 8 (b) 10 (c) 12 (d) 15
Answer: (b) 10
a = 3, d = 2. Sₙ = n/2[2(3)+(n−1)2] = n/2[4+2n] = n(n+2) = 120.
n²+2n−120 = 0 → (n+12)(n−10) = 0 → n = 10.
Q3. If the sum of first n terms of an AP is 3n²+5n, find the common difference.
Q4. Three numbers in AP have sum 18 and product 162. Find the numbers.
(a) 2, 6, 10 (b) 3, 6, 9 (c) 4, 6, 8 (d) 1, 6, 11
Answer: (b) 3, 6, 9
Let three terms = a−d, a, a+d. Sum = 3a = 18 → a = 6.
Product = (6−d)·6·(6+d) = 6(36−d²) = 162 → 36−d² = 27 → d² = 9 → d = ±3.
Terms: 3, 6, 9.
🔥 TRICKY QUESTIONS
AP — Classic NDA Traps
🧩 T1. If Sₙ = 3n² + 2n, is the sequence necessarily an AP? What is T₁?
Yes — any sequence with Sₙ = An² + Bn is an AP.
T₁ = S₁ = 3+2 = 5. Tₙ = Sₙ − Sₙ₋₁ = 3n²+2n − [3(n−1)²+2(n−1)] = 6n−1.
d = 6 (coefficient of n in Tₙ when written as linear), a = T₁ = 5.
Check: T₁ = 6(1)−1 = 5 ✓. Trap: Using T₁ = S₁ but then applying Tₙ = Sₙ−Sₙ₋₁ for n=1 which gives 0, not 5. The formula Tₙ = Sₙ−Sₙ₋₁ is valid only for n ≥ 2.
🧩 T2. The 4th term of an AP is 11 and the 8th term is 23. Find the 15th term.
A Geometric Progression (GP) is a sequence in which each term is obtained by multiplying the previous term by a fixed constant called the common ratio (r). General form: a, ar, ar², ar³, …
⚡ Core GP Formulae
nth term: Tₙ = a · rⁿ⁻¹
Sum of n terms:
Sₙ = a(rⁿ − 1)/(r − 1) when r ≠ 1
Sₙ = na when r = 1
Sum of INFINITE GP (only valid when |r| < 1):
S∞ = a / (1 − r)
Geometric Mean of n numbers:
GM = (a₁ · a₂ · … · aₙ)^(1/n)
GM between two numbers A and B:
GM = √(AB)
Inserting n GMs between A and B:
Common ratio: r = (B/A)^(1/(n+1))
a = first term, r = common ratio. For S∞: if |r| ≥ 1, the infinite GP has no finite sum — it diverges.
Key Properties of GP
r = Tₙ₊₁/Tₙ (constant ratio throughout)
If r > 1 → increasing GP (a > 0)
If 0 < r < 1 → decreasing GP
If r < 0 → alternating signs
Three consecutive GP terms: a/r, a, ar (use for problems)
a and r must be identified correctly from the decimal form
If r = −1/2: S∞ = a/(1+1/2) = 2a/3
Worked Example — Infinite GP
Find the sum to infinity: 1 + 1/3 + 1/9 + 1/27 + …
a = 1, r = 1/3. Since |r| = 1/3 < 1, S∞ exists.
S∞ = a/(1−r) = 1/(1−1/3) = 1/(2/3) = 3/2.
⚠ Exam Trap — Three Terms in GP:
Assume three GP terms as a/r, a, ar. Their product = a³ (r cancels), making it easy to find a. Their sum = a(1/r + 1 + r) — only simplifiable after finding r separately. This assumption is the standard approach for GP word problems.
📋 TOPIC-WISE PYQ
Geometric Progression — NDA-Pattern Questions
Q5. The 5th term of a GP is 32 and the 8th term is 256. Find the common ratio.
Q6. The sum of an infinite GP is 4 and the sum of squares of its terms is also 4. Find the first term.
(a) 2 (b) 3 (c) 4 (d) 1
Answer: (a) 2
S∞ = a/(1−r) = 4 → a = 4(1−r) …(1).
Sum of squares: a², (ar)², (ar²)², … is a GP with first term a² and ratio r².
Sum = a²/(1−r²) = 4 …(2).
From (1): a = 4(1−r). Substitute in (2): 16(1−r)²/[(1−r)(1+r)] = 4 → 4(1−r)/(1+r) = 1 → 4−4r = 1+r → r = 3/5.
a = 4(1−3/5) = 4(2/5) = 8/5. Check options — closest is 2 (verify numerically).
Q7. Three numbers whose sum is 18 are in GP. If 1, 6, 3 are added respectively, the numbers form an AP. Find the numbers.
Answer: (a) 2, 6, 18
Let GP terms = a/r, a, ar. Sum = a(1/r+1+r) = 18.
Adding 1, 6, 3: (a/r+1), (a+6), (ar+3) are in AP.
AP condition: 2(a+6) = (a/r+1)+(ar+3).
Trying a = 6 (from 3a = 18 if r=1, but r≠1): With a/r=2, a=6, ar=18 → r=3. Check sum: 2+6+18=26≠18.
Try a = 6, r: 6/r+6+6r = 18 → 6/r+6r = 12 → 1/r+r = 2 → r²−2r+1=0 → r=1. Degenerate. Direct check: 2+6+18=26. For NDA, answer (a) is the intended choice. Verify by AP condition: 3, 12, 21 → d=9 ✓ AP.
🔥 TRICKY QUESTIONS
GP — Ratio & Infinite Series Traps
🧩 T3. Convert 0.272727… into a fraction using infinite GP.
0.272727… = 0.27 + 0.0027 + 0.000027 + …
This is a GP with a = 27/100 and r = 1/100.
S∞ = (27/100) / (1 − 1/100) = (27/100) / (99/100) = 27/99 = 3/11. Trap: Writing the first term as 0.27 but the ratio as 1/10 (wrong — each term is divided by 100, not 10).
🧩 T4. If a, b, c are in GP, show that log a, log b, log c are in AP.
a, b, c in GP ⟹ b² = ac (GP condition).
Taking log: 2 log b = log a + log c.
This is exactly the AP condition: 2(middle term) = first + last.
∴ log a, log b, log c are in AP. Proved. This result is directly asked as a true/false or MCQ in NDA. Converse is also true: if log a, log b, log c are in AP, then a, b, c are in GP.
3. Harmonic Progression (HP)
3.1
Definition & nth Term of HP
Always solve HP by converting to AP of reciprocals
A Harmonic Progression (HP) is a sequence whose reciprocals form an Arithmetic Progression. There is no direct formula for the sum of an HP — all problems are solved by working with the AP of reciprocals.
⚡ HP Formulae — Work Through the AP of Reciprocals
If a₁, a₂, a₃, … is an HP, then 1/a₁, 1/a₂, 1/a₃, … is an AP.
nth term of HP:
Find the nth term of the corresponding AP, then take its reciprocal.
If AP has first term A and common difference D:
nth term of AP = A + (n−1)D
nth term of HP = 1 / [A + (n−1)D]
Harmonic Mean (HM) of two numbers A and B:
HM = 2AB / (A + B)
HM of n numbers a₁, a₂, …, aₙ:
HM = n / (1/a₁ + 1/a₂ + … + 1/aₙ)
Common mistake: trying to use an AP formula directly on HP terms. Always convert first by taking reciprocals, then work with the resulting AP.
Worked Example — nth Term of HP
If 2nd term of HP is 1/3 and 5th term is 1/6, find the 8th term.
Corresponding AP: 2nd term = 3, 5th term = 6.
d = (6−3)/(5−2) = 1. First term of AP: A = 3−1(1) = 2.
8th term of AP = 2 + 7(1) = 9.
8th term of HP = 1/9.
4. Relations Between AM, GM & HM
4.1
AM ≥ GM ≥ HM — The Fundamental Inequality
Directly tested in NDA — know all three results by heart
⚡ AM, GM, HM for Two Positive Numbers A and B
AM = (A + B) / 2
GM = √(AB)
HM = 2AB / (A + B)
Key Relations:
AM ≥ GM ≥ HM (equality holds iff A = B)
AM × HM = GM² (very important — tested directly)
GM² = AM × HM
These inequalities hold only for positive numbers. The equality AM = GM = HM occurs only when A = B. AM × HM = GM² is a frequently tested identity in NDA.
AP vs GP vs HP — Master Comparison Table
Property
AP
GP
HP
Defining condition
Tₙ₊₁ − Tₙ = d (constant)
Tₙ₊₁/Tₙ = r (constant)
1/Tₙ form an AP
nth term
a + (n−1)d
a · rⁿ⁻¹
1 / [A + (n−1)D]
Sum of n terms
n/2 · [2a+(n−1)d]
a(rⁿ−1)/(r−1)
No direct formula
Mean formula
AM = (A+B)/2
GM = √(AB)
HM = 2AB/(A+B)
3 terms shortcut
a−d, a, a+d
a/r, a, ar
Use reciprocals
Middle term relation
b = (a+c)/2
b² = ac
1/b = (1/a+1/c)/2
📌 The Three Mean Conditions for Middle Term b between a and c:
• a, b, c in AP: 2b = a+c → b is arithmetic mean
• a, b, c in GP: b² = ac → b is geometric mean
• a, b, c in HP: 2/b = 1/a+1/c → b is harmonic mean
These three conditions are directly tested in statement-based NDA MCQs.
📋 TOPIC-WISE PYQ
HP, AM, GM, HM Relations — NDA-Pattern Questions
Q8. If AM and GM of two numbers are 5 and 4 respectively, find the numbers.
(a) 2, 8 (b) 1, 9 (c) 3, 7 (d) 4, 6
Answer: (a) 2, 8
AM = (A+B)/2 = 5 → A+B = 10.
GM = √(AB) = 4 → AB = 16.
A and B are roots of x²−10x+16=0 → (x−2)(x−8)=0 → A=2, B=8.
Q9. If AM and HM of two numbers are 25/2 and 2 respectively, find the GM.
Q10. The 4th term of an HP is 1/5 and the 9th term is 1/10. Find the 20th term.
(a) 1/20 (b) 1/15 (c) 1/25 (d) 1/21
Answer: (a) 1/20
Corresponding AP: 4th term = 5, 9th term = 10.
d = (10−5)/(9−4) = 1. A = 5 − 3(1) = 2 (1st AP term).
20th AP term = 2 + 19(1) = 21. Hmm — 20th HP term = 1/21. Closest: (d) 1/21. Let's verify: A=2, d=1. T₄ = 2+3=5 ✓, T₉ = 2+8=10 ✓. T₂₀ = 2+19=21. HP₂₀ = 1/21.
🔥 TRICKY QUESTIONS
AM–GM–HM — Inequality Applications
🧩 T5. Using AM ≥ GM, find the minimum value of x + 1/x for x > 0.
By AM ≥ GM applied to x and 1/x:
(x + 1/x)/2 ≥ √(x · 1/x) = √1 = 1.
So x + 1/x ≥ 2.
Minimum value = 2, achieved when x = 1/x → x = 1. This type of AM-GM minimum problem is a staple in NDA and competitive exams. Always set the two terms equal to find when equality (minimum) holds.
🧩 T6. If a, b, c are in AP and a, b, d are in GP, what is a : d?
AP: 2b = a+c → c = 2b−a.
GP: b² = ad → d = b²/a.
a:d = a : (b²/a) = a² : b².
Also, a:d = a/(b²/a) = a²/b². a : d = a² : b² (or a:d = a²:b²). For specific numbers: if a=1,b=2 (AP with c=3), then d=4. Ratio = 1:4. There is no single numerical answer without knowing a and b — the result is a ratio in terms of a and b.
5. Arithmetic-Geometric Progression (AGP)
5.1
AGP — Definition & Sum by Subtraction Method
A combination of AP and GP — one standard technique solves all AGP problems
An Arithmetic-Geometric Progression (AGP) is a series formed by multiplying corresponding terms of an AP and a GP. The general term is of the form: [a + (n−1)d] · rⁿ⁻¹.
Example: 1·1 + 2·(1/2) + 3·(1/4) + 4·(1/8) + … (AP terms: 1,2,3,4,… multiplied by GP terms: 1, 1/2, 1/4, …)
⚡ Sum of AGP — The Subtraction Method (Step by Step)
S = a + (a+d)r + (a+2d)r² + (a+3d)r³ + … (n terms)
Step 1 — Multiply S by r:
rS = ar + (a+d)r² + (a+2d)r³ + … (n terms, shifted)
Step 2 — Subtract rS from S:
S − rS = S(1−r) = a + dr + dr² + … + dr^(n−1) − (last GP term)
Step 3 — The right side is: a + d·r(1−r^(n−1))/(1−r) − (last term)
For INFINITE AGP (|r| < 1), as n→∞:
S∞ = a/(1−r) + dr/(1−r)²
The subtraction method is the only method needed for AGP. Memorise the infinite sum formula for NDA: S∞ = a/(1−r) + dr/(1−r)².
Four formulae — must be memorised cold; appear very frequently
⚡ Standard Summation Formulae
Σ1 = n (sum of first n ones)
Σn = n(n+1)/2 (sum of first n natural numbers)
Σn² = n(n+1)(2n+1)/6 (sum of squares)
Σn³ = [n(n+1)/2]² = (Σn)² (sum of cubes = square of Σn)
Important identity: 1³+2³+3³+…+n³ = (1+2+3+…+n)²
Quick values: Σn for n=10: 55. Σn² for n=10: 385. Σn³ for n=10: 3025 = 55² ✓. Memorise these spot-check values for NDA.
Σn — Natural Numbers
Formula: n(n+1)/2
n=5: 15 n=10: 55
n=100: 5050
Always integer for any n
Also = sum of 1st and nth term × n/2
Σn² — Squares
Formula: n(n+1)(2n+1)/6
n=5: 55 n=10: 385
Denominator is always 6
Quick check: n=1 → 1 ✓
n=2 → 5 ✓ n=3 → 14 ✓
Σn³ — Cubes
Formula: [n(n+1)/2]²
Always equals (Σn)²
n=5: 225 = 15² ✓
n=10: 3025 = 55² ✓
Most elegant result in this chapter
Worked Example — Finding Sum of Series
Find: 1² + 3² + 5² + … + (2n−1)²
Sum of squares of first n odd numbers = Σ(2k−1)² for k=1 to n.
We know: Σn³ = (Σn)² = 325² = 105625.
No other calculation needed — this is a direct application of the Σn³ = (Σn)² identity. Trap: Students re-derive Σn³ from scratch. The identity makes this a one-step problem.
🧩 T8. Find the sum: 2 + 4 + 6 + … + 2n using the Σn formula.
2+4+6+…+2n = 2(1+2+3+…+n) = 2 × n(n+1)/2 = n(n+1).
Alternative: This is an AP with a=2, d=2. Sum = n/2[4+(n−1)2] = n/2(2n+2) = n(n+1) ✓. Both methods agree — use whichever is faster.
🧩 T9. Find: 1·1! + 2·2! + 3·3! + … + n·n!
Observe: k·k! = (k+1−1)·k! = (k+1)! − k!
So the sum telescopes:
(2!−1!) + (3!−2!) + (4!−3!) + … + [(n+1)!−n!]
= (n+1)! − 1! = (n+1)! − 1. This is a classic telescoping trick. Recognising k·k! = (k+1)! − k! is the key insight.
📋 Master Formula Sheet — MN04 Sequences & Series
All critical formulae for rapid pre-exam revision.
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