Sequences and Series

Sequences and Series

Board: CBSE |Class:Class11

ComprehensivestudynotesforSequencesandSeriesbyAjayYadav(MathKingofKatargam).Mastereveryconceptwithclearexplanations,solvedexamples,andpracticeproblems.

KeyConceptsArithmeticProgression(AP)

Sequencewhereeachtermdiffersbyconstantd.nthterm:a₅=a+(n-1)d.Sum:S₅=n/2[2a+(n-1)d]=n/2(a+l).

GeometricProgression(GP)

Sequencewhereeachtermismultipliedbyconstantr.nthterm:a₅=arⁿ⁻¹.Sum(r≠1):S₅=a(rⁿ-1)/(r-1).SumofinfiniteGP(|r|<1): S∞ = a/(1-r).

Harmonic Progression (HP)

Sequence whose reciprocals form an AP. Example: 1/a, 1/(a+d), 1/(a+2d), …

Arithmetic Mean

AM of a and b = (a+b)/2. If a, A, b are in AP, then A is the AM.

Geometric Mean

GM of a and b = ab. If a, G, b are in GP, then G is the GM.

Relationship between AM and GM

AM ≥ GM for positive numbers. AM = GM iff all numbers are equal. For n numbers: (a1+…+aₙ)/n ≥ (a1…aₙ)¹/ⁿ.

Sum of Special Series

Sum of first n natural numbers: Σn = n(n+1)/2. Sum of squares: Σn² = n(n+1)(2n+1)/6. Sum of cubes: Σn³ = [n(n+1)/2]².

Important Formulas

AP nth terma₅ = a + (n-1)d
AP SumS₅ = n/2[2a+(n-1)d]
GP nth terma₅ = arⁿ⁻¹
GP Sum (finite)S₅ = a(rⁿ-1)/(r-1)
GP Sum (infinite)S∞ = a/(1-r)
Σnn(n+1)/2
Σn²n(n+1)(2n+1)/6

Solved Examples

Example 1: Find sum of first 20 terms of AP: 2, 5, 8, …

Solution: a=2, d=3, n=20. S = 20/2[4+19(3)] = 10(61) = 610.

Example 2: Find sum of GP: 1, 1/2, 1/4, … infinite.

Solution: a=1, r=1/2. S∞ = 1/(1-1/2) = 2.

Example 3: Find Σn² for n=1 to 10.

Solution: 10(11)(21)/6 = 385.

Practice Questions

  1. Find the 10th term of AP: 1, 4, 7, 10, …
  2. Find the sum of GP: 2, 4, 8, 16, … to 10 terms.
  3. Insert 3 AMs between 2 and 10.
  4. The sum of three numbers in GP is 21 and product is 216. Find the numbers.
  5. Find the sum: 1² + 2² + 3² + … + 20².

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