1. Sequence and Series 

Sequence: A set of numbers arranged according to a rule.

Can be finite or infinite.

Examples:

2, 4, 6, 8, … (infinite)

5, 10, 15, 20, 25 (finite)

Series: Sum of the terms of a sequence.

Example: Sequence 2, 4, 6, 8 → Series = 2 + 4 + 6 + 8 = 20

Types of Sequences:

Arithmetic Sequence (AP): Difference between consecutive terms is constant.

Geometric Sequence (GP): Ratio between consecutive terms is constant.

Fibonacci Sequence: Each term is sum of previous two terms. Example: 0, 1, 1, 2, 3, 5, 8, …

Harmonic Sequence (HP): Reciprocal of an AP. Example: 1, 1/2, 1/3, 1/4, …

2. Arithmetic Progression (AP)

Key Points:

Constant difference = d = t₂ − t₁

tₙ = a + (n − 1)d

Sum of n terms: Sₙ = n/2 × (2a + (n − 1)d) or Sₙ = n/2 × (a + l)

Properties of AP:

Middle term of odd number of terms = arithmetic mean of first and last term

Sum of terms equidistant from beginning and end = same

Example: AP = 2, 5, 8, 11, 14 → 2 + 14 = 16, 5 + 11 = 16

3. Geometric Progression (GP)

Key Points:

Constant ratio = r = t₂ / t₁

tₙ = a × rⁿ⁻¹

Sum of n terms: Sₙ = a × (rⁿ − 1)/(r − 1) (r ≠ 1)

Special Cases:

Sum of infinite GP (|r| < 1) → S∞ = a / (1 − r)

Product of n terms → t₁ × t₂ × … × tₙ = aⁿ × r^(n(n−1)/2)

Properties of GP:

Any term squared = product of terms equidistant from beginning and end

Example: GP = 3, 6, 12, 24, 48 → 12² = 6 × 24

4. Important Solved Questions – Extra Practice

Q1. Find 30th term of AP: 7, 12, 17, …

a = 7, d = 5

t₃₀ = a + (n − 1)d

t₃₀ = 7 + (30 − 1) × 5

t₃₀ = 7 + 29 × 5

t₃₀ = 7 + 145

t₃₀ = 152

Q2. Sum of first 20 terms of AP: 3, 8, 13, …

a = 3, d = 5, n = 20

Sₙ = n/2 × [2a + (n − 1)d]

S₂₀ = 20/2 × [2×3 + (20 − 1)×5]

S₂₀ = 10 × [6 + 95]

S₂₀ = 10 × 101

S₂₀ = 1010

Q3. Insert 4 AMs between 10 and 50

Number of AMs = 4 → divide interval by (n + 1) = 5

d = (50 − 10) / 5

d = 40 / 5 = 8

AMs = a + d, a + 2d, a + 3d, a + 4d

AMs = 10 + 8 = 18, 10 + 16 = 26, 10 + 24 = 34, 10 + 32 = 42

AMs = 18, 26, 34, 42

Q4. Find 6th term of GP: 2, 6, 18, …

a = 2, r = 3, n = 6

tₙ = a × rⁿ⁻¹

t₆ = 2 × 3⁵

t₆ = 2 × 243

t₆ = 486

Q5. Sum of first 5 terms of GP: 1, 2, 4, 8, 16

a = 1, r = 2, n = 5

Sₙ = a × (rⁿ − 1) / (r − 1)

S₅ = 1 × (2⁵ − 1) / (2 − 1)

S₅ = (32 − 1) / 1

S₅ = 31

Q6. GP with t₃ = 9 and t₆ = 72 → Find a, r, and sequence

t₃ = a × r² = 9

t₆ = a × r⁵ = 72

Step 1: Divide t₆ by t₃ → r³ = 72 / 9 = 8 → r = 2

Step 2: Solve for a → t₃ = a × r² = 9 → a × 4 = 9 → a = 9/4 = 2.25

Step 3: Sequence → t₁ = 2.25, t₂ = 4.5, t₃ = 9, t₄ = 18, t₅ = 36, t₆ = 72

Sequence = 2.25, 4.5, 9, 18, 36, 72

Q7. Find sum of first 10 natural numbers using AP formula

a = 1, d = 1, n = 10

Sₙ = n/2 × [2a + (n − 1)d]

S₁₀ = 10/2 × [2×1 + (10 − 1)×1]

S₁₀ = 5 × [2 + 9]

S₁₀ = 5 × 11

S₁₀ = 55

5. Extra Challenging Questions – For Practice

Find 50th term of AP: 2, 5, 8, …

Sum of first 25 terms of AP: 7, 10, 13, …

Insert 7 AMs between −3 and 25

Find 12th term of GP: 1, 3, 9, …

Sum of first 6 terms of GP: 2, 4, 8, 16, 32, 64

GP with t₂ = 6, t₅ = 48 → Find first term and sequence

If sum of first n terms of AP = 210, a = 3, d = 5 → Find n

If sum of first n terms of GP = 364, a = 1, r = 2 → Find n

6. Tips for Understanding AP and GP

Draw sequences on number lines to visualize progression

Check common difference / ratio first before solving

For sums, always identify first term (a) and last term (l) or use formula

For inserting AMs, divide the gap by n+1

For challenging GP problems, use division method to find ratio r.

Visit this link for further practice !!

https://besidedegree.com/exam/s/academic