Concept Core
AP, GP, AM-GM inequality, and special series — the complete sequences framework.
Arithmetic Progression (AP)
A sequence a, a+d, a+2d, ... with constant common difference d.
nth term: aₙ = a + (n−1)d
Sum of n terms: Sₙ = n/2 · (2a + (n−1)d) = n/2 · (a + l)
Sum of first n natural numbers: n(n+1)/2
Sum of first n odd numbers: n²
Arithmetic Mean (AM) of a and b: AM = (a+b)/2. If a, A, b are in AP: A = (a+b)/2.
Geometric Progression (GP)
A sequence a, ar, ar², ... with constant common ratio r.
nth term: aₙ = arⁿ⁻¹
Sum of n terms: Sₙ = a(rⁿ−1)/(r−1) for r ≠ 1
Sum of infinite GP (|r|<1): S∞ = a/(1−r)
Geometric Mean: GM = √(ab)
Harmonic Progression (HP)
A sequence whose reciprocals are in AP. If a, b, c are in HP:
1/a, 1/b, 1/c are in AP → 2/b = 1/a + 1/c
Harmonic Mean: HM = 2ab/(a+b)
AM–GM–HM inequality: AM ≥ GM ≥ HM (equality iff all terms equal)
Special Series Sums
Σn = n(n+1)/2
Σn² = n(n+1)(2n+1)/6
Σn³ = [n(n+1)/2]² = (Σn)²
Σ(2n−1) = n² (sum of n odd numbers)
Inserting Means Between Two Numbers
n arithmetic means between a and b: d = (b−a)/(n+1). Means are a+d, a+2d, ..., a+nd.
n geometric means between a and b: r = (b/a)^(1/(n+1)). Means are ar, ar², ..., arⁿ.
Method of Differences
For a sequence where differences follow a pattern, find the nth term by the difference method:
If 1st differences form AP → Tₙ is quadratic. If 1st differences form GP → use sum of GP.
Sum by vⁿ method: split Tₙ = f(n)−f(n−1) and telescope the sum.
Formula Vault
All sequence and series formulas for EAPCET.
AP nth Term
aₙ = a + (n−1)d
a = first term; d = common difference
AP Sum
Sₙ = n/2·(2a+(n−1)d)
Also = n/2·(a+l), l = last term
GP nth Term
aₙ = arⁿ⁻¹
r = common ratio
GP Sum (finite)
Sₙ = a(rⁿ−1)/(r−1)
For r ≠ 1
Infinite GP
S∞ = a/(1−r)
Valid only when |r| < 1
AM between a,b
AM = (a+b)/2
AM ≥ GM ≥ HM
GM between a,b
GM = √(ab)
For positive a, b
HM between a,b
HM = 2ab/(a+b)
Reciprocals in AP
Sum of n²
Σn² = n(n+1)(2n+1)/6
Sum of squares of first n naturals
Sum of n³
Σn³ = [n(n+1)/2]²
= (Σn)² — beautiful identity
Worked Examples
5 problems — AP/GP terms, sums, infinite GP, AM-GM, and a common trap.
EasyFind the 20th term of the AP: 3, 7, 11, 15, ...▾
Find the 20th term of the AP 3, 7, 11, 15, ...
2
a₂₀ = a + 19d = 3 + 19 × 4 = 3 + 76 = 79
✓ a₂₀ = 79
EasySum of the first 10 terms of GP: 2, 6, 18, ...▾
Find the sum of the first 10 terms of the GP 2, 6, 18, ...
2
S₁₀ = a(r¹⁰−1)/(r−1) = 2(3¹⁰−1)/(3−1) = 2(59049−1)/2 = 59048
✓ S₁₀ = 59048
MediumFind the sum: 0.5 + 0.05 + 0.005 + ... (infinite GP)▾
Find the sum of the infinite series 0.5 + 0.05 + 0.005 + ...
1
a = 0.5, r = 0.1 (|r| < 1, so infinite sum exists)
2
S∞ = a/(1−r) = 0.5/(1−0.1) = 0.5/0.9 = 5/9
✓ S∞ = 5/9
EAPCET LevelIf AM = 10 and GM = 8 between two positive numbers, find HM and the numbers▾
Two positive numbers have AM = 10 and GM = 8. Find HM and the two numbers.
1
HM = GM²/AM = 64/10 = 6.4 (using AM·HM = GM²)
2
Numbers: a + b = 2×AM = 20; ab = GM² = 64
3
So a and b satisfy: t² − 20t + 64 = 0
4
t = (20 ± √(400−256))/2 = (20 ± 12)/2
✓ HM = 6.4; Numbers are 16 and 4
Trap QuestionThe sum of an infinite GP always exists — True or False?▾
Is it true that every GP with positive terms has a finite infinite sum?
1
The trap: The infinite sum S∞ = a/(1−r) only exists when |r| < 1.
2
For the GP 2, 4, 8, 16, ... → r = 2 > 1. The series diverges — no finite sum.
3
For r = 1: all terms equal a, sum is infinite.
4
For r = −1: terms alternate, partial sums oscillate, no limit.
5
Only |r| < 1 guarantees convergence.
✓ False — infinite GP sum exists only when |r| < 1
Mistake DNA
4 sequence and series errors from EAPCET distractor analysis.
🔢
Using AP Sum Formula for GP and Vice Versa
The two formulas are completely different. Mixing them up is the most common error.
❌ Wrong
GP sum: Sₙ = n/2·(2a+(n−1)d) ✗
(That's AP formula!)
✓ Correct
AP: Sₙ = n/2·(2a+(n−1)d) ✓
GP: Sₙ = a(rⁿ−1)/(r−1) ✓
Identify which type first
First question to ask: is the difference constant (AP) or the ratio constant (GP)? Write out 2-3 terms to verify before applying any formula.
∞
Applying Infinite GP Formula When |r| ≥ 1
S∞ = a/(1−r) requires |r| < 1. For |r| ≥ 1 the series diverges — there is no finite sum.
❌ Wrong
GP: 3, 6, 12, ... (r=2)
S∞ = 3/(1−2) = −3 ✗
(r > 1: no finite sum)
✓ Correct
|r| < 1 required for S∞ ✓
r = 2 > 1: series diverges ✓
No finite infinite sum exists
Always check |r| < 1 before using the infinite GP formula. If |r| ≥ 1, the infinite sum does not exist and the formula gives a meaningless or negative answer.
📐
AM–GM: GM² = AM × HM (Forgetting This Identity)
The beautiful identity AM × HM = GM² is frequently tested. Students compute HM from scratch instead.
❌ Wrong
Find HM given AM=9, GM=6:
HM = 2ab/(a+b) (compute
a,b first — long method)
✓ Correct
AM × HM = GM² ✓
HM = GM²/AM = 36/9 = 4 ✓
Two steps, no need to
find a and b
AM × HM = GM² is a direct identity. Given any two of AM, GM, HM, find the third instantly. Don't solve for the original numbers unless explicitly asked.
🔑
nth Term vs Sum Formula Confusion
aₙ = a + (n−1)d gives the nth TERM. Sₙ gives the SUM of n terms. They are different.
❌ Wrong
'Find 10th term of AP':
S₁₀ = 10/2·(2a+9d) ✗
(That's sum of 10 terms!)
✓ Correct
10th term: a₁₀ = a + 9d ✓
'Sum of first 10 terms':
S₁₀ = 10/2·(2a+9d) ✓
Read the question carefully
If asked for 'the nth term': use aₙ = a+(n−1)d. If asked for 'sum of first n terms': use Sₙ formula. Also: Tₙ = Sₙ − Sₙ₋₁ (useful when only Sₙ is given).
Chapter Intelligence
Sequences & Series is a direct-marks chapter — formulas here are quick to apply.
EAPCET Weightage (2019–2024)
Special series Σn, Σn², Σn³~4 High-Yield PYQ Patterns
Find nth term of AP/GPSum of n terms of AP/GPAM·HM = GM² problemsInfinite GP: find sumInsert n means between two numbersΣn² and Σn³ direct computationThree numbers in AP/GP: find them
Exam Strategy
- Three numbers in AP: take them as (a−d), a, (a+d). Three numbers in GP: take as a/r, a, ar. This simplifies the algebra enormously.
- Infinite GP: first verify |r| < 1, then S∞ = a/(1−r). Never apply this formula without checking the ratio.
- AM × HM = GM². This identity solves half of all AM-GM-HM problems in one step. Commit it to memory.
- Sₙ = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]². These appear as direct substitution MCQs every paper.