Concept Core
AP, GP, AM-GM inequality, and special series β the complete sequences framework.
Arithmetic Progression (AP)
A sequence a, a+d, a+2d, ... with constant common difference d.
nth term: aβ = a + (nβ1)d
Sum of n terms: Sβ = n/2 Β· (2a + (nβ1)d) = n/2 Β· (a + l)
Sum of first n natural numbers: n(n+1)/2
Sum of first n odd numbers: nΒ²
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, arΒ², ... with constant common ratio r.
nth term: aβ = arβΏβ»ΒΉ
Sum of n terms: Sβ = a(rβΏβ1)/(rβ1) for r β 1
Sum of infinite GP (|r|<1): Sβ = a/(1βr)
Geometric Mean: GM = β(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 β 2/b = 1/a + 1/c
Harmonic Mean: HM = 2ab/(a+b)
AMβGMβHM inequality: AM β₯ GM β₯ HM (equality iff all terms equal)
Special Series Sums
Ξ£n = n(n+1)/2
Ξ£nΒ² = n(n+1)(2n+1)/6
Ξ£nΒ³ = [n(n+1)/2]Β² = (Ξ£n)Β²
Ξ£(2nβ1) = nΒ² (sum of n odd numbers)
Inserting Means Between Two Numbers
n arithmetic means between a and b: d = (bβa)/(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, arΒ², ..., arβΏ.
Method of Differences
For a sequence where differences follow a pattern, find the nth term by the difference method:
If 1st differences form AP β Tβ is quadratic. If 1st differences form GP β use sum of GP.
Sum by vβΏ method: split Tβ = f(n)βf(nβ1) and telescope the sum.
Formula Vault
All sequence and series formulas for EAPCET.
AP nth Term
aβ = a + (nβ1)d
a = first term; d = common difference
AP Sum
Sβ = n/2Β·(2a+(nβ1)d)
Also = n/2Β·(a+l), l = last term
GP nth Term
aβ = arβΏβ»ΒΉ
r = common ratio
GP Sum (finite)
Sβ = a(rβΏβ1)/(rβ1)
For r β 1
Infinite GP
Sβ = a/(1βr)
Valid only when |r| < 1
AM between a,b
AM = (a+b)/2
AM β₯ GM β₯ HM
GM between a,b
GM = β(ab)
For positive a, b
HM between a,b
HM = 2ab/(a+b)
Reciprocals in AP
Sum of nΒ²
Ξ£nΒ² = n(n+1)(2n+1)/6
Sum of squares of first n naturals
Sum of nΒ³
Ξ£nΒ³ = [n(n+1)/2]Β²
= (Ξ£n)Β² β beautiful identity
Worked Examples
5 problems β AP/GP terms, sums, infinite GP, AM-GM, and a common trap.
EasyFind the 20th term of the AP: 3, 7, 11, 15, ...βΎ
Find the 20th term of the AP 3, 7, 11, 15, ...
2
aββ = a + 19d = 3 + 19 Γ 4 = 3 + 76 = 79
β aββ = 79
EasySum of the first 10 terms of GP: 2, 6, 18, ...βΎ
Find the sum of the first 10 terms of the GP 2, 6, 18, ...
2
Sββ = a(rΒΉβ°β1)/(rβ1) = 2(3ΒΉβ°β1)/(3β1) = 2(59049β1)/2 = 59048
β Sββ = 59048
MediumFind the sum: 0.5 + 0.05 + 0.005 + ... (infinite GP)βΎ
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
Sβ = a/(1βr) = 0.5/(1β0.1) = 0.5/0.9 = 5/9
β Sβ = 5/9
EAPCET LevelIf AM = 10 and GM = 8 between two positive numbers, find HM and the numbersβΎ
Two positive numbers have AM = 10 and GM = 8. Find HM and the two numbers.
1
HM = GMΒ²/AM = 64/10 = 6.4 (using AMΒ·HM = GMΒ²)
2
Numbers: a + b = 2ΓAM = 20; ab = GMΒ² = 64
3
So a and b satisfy: tΒ² β 20t + 64 = 0
4
t = (20 Β± β(400β256))/2 = (20 Β± 12)/2
β HM = 6.4; Numbers are 16 and 4
Trap QuestionThe sum of an infinite GP always exists β True or False?βΎ
Is it true that every GP with positive terms has a finite infinite sum?
1
The trap: The infinite sum Sβ = a/(1βr) only exists when |r| < 1.
2
For the GP 2, 4, 8, 16, ... β r = 2 > 1. The series diverges β no finite sum.
3
For r = 1: all terms equal a, sum is infinite.
4
For r = β1: terms alternate, partial sums oscillate, no limit.
5
Only |r| < 1 guarantees convergence.
β False β infinite GP sum exists only when |r| < 1
Mistake DNA
4 sequence and series errors from EAPCET distractor analysis.
π’
Using AP Sum Formula for GP and Vice Versa
The two formulas are completely different. Mixing them up is the most common error.
β Wrong
GP sum: Sβ = n/2Β·(2a+(nβ1)d) β
(That's AP formula!)
β Correct
AP: Sβ = n/2Β·(2a+(nβ1)d) β
GP: Sβ = a(rβΏβ1)/(rβ1) β
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.
β
Applying Infinite GP Formula When |r| β₯ 1
Sβ = a/(1βr) requires |r| < 1. For |r| β₯ 1 the series diverges β there is no finite sum.
β Wrong
GP: 3, 6, 12, ... (r=2)
Sβ = 3/(1β2) = β3 β
(r > 1: no finite sum)
β Correct
|r| < 1 required for Sβ β
r = 2 > 1: series diverges β
No finite infinite sum exists
Always check |r| < 1 before using the infinite GP formula. If |r| β₯ 1, the infinite sum does not exist and the formula gives a meaningless or negative answer.
π
AMβGM: GMΒ² = AM Γ HM (Forgetting This Identity)
The beautiful identity AM Γ HM = GMΒ² is frequently tested. Students compute HM from scratch instead.
β Wrong
Find HM given AM=9, GM=6:
HM = 2ab/(a+b) (compute
a,b first β long method)
β Correct
AM Γ HM = GMΒ² β
HM = GMΒ²/AM = 36/9 = 4 β
Two steps, no need to
find a and b
AM Γ HM = GMΒ² 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.
π
nth Term vs Sum Formula Confusion
aβ = a + (nβ1)d gives the nth TERM. Sβ gives the SUM of n terms. They are different.
β Wrong
'Find 10th term of AP':
Sββ = 10/2Β·(2a+9d) β
(That's sum of 10 terms!)
β Correct
10th term: aββ = a + 9d β
'Sum of first 10 terms':
Sββ = 10/2Β·(2a+9d) β
Read the question carefully
If asked for 'the nth term': use aβ = a+(nβ1)d. If asked for 'sum of first n terms': use Sβ formula. Also: Tβ = Sβ β Sβββ (useful when only Sβ is given).
Chapter Intelligence
Sequences & Series is a direct-marks chapter β formulas here are quick to apply.
EAPCET Weightage (2019β2024)
Special series Ξ£n, Ξ£nΒ², Ξ£nΒ³~4 High-Yield PYQ Patterns
Find nth term of AP/GPSum of n terms of AP/GPAMΒ·HM = GMΒ² problemsInfinite GP: find sumInsert n means between two numbersΞ£nΒ² and Ξ£nΒ³ direct computationThree numbers in AP/GP: find them
Exam Strategy
- Three numbers in AP: take them as (aβd), 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 Sβ = a/(1βr). Never apply this formula without checking the ratio.
- AM Γ HM = GMΒ². This identity solves half of all AM-GM-HM problems in one step. Commit it to memory.
- Sβ = n(n+1)/2, Ξ£nΒ² = n(n+1)(2n+1)/6, Ξ£nΒ³ = [n(n+1)/2]Β². These appear as direct substitution MCQs every paper.
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