1. Sequence and Series
1.1 Sequence: A set of numbers arranged according to a rule.
Types:
1.Finite sequence
2.Infinite sequence
Examples:
1.2,4,6,8,… (infinite)
2.5,10,15,20,25 (finite)
1.2 Series
A series is the sum of the terms of a sequence.
Example:
Sequence:2,4,6,8
Series:2+4+6+8=20
1.3Types of Sequences
1.Arithmetic Progression(AP)
•Constant difference between consecutive terms
2.Geometric Progression(GP)
•Constant ratio between consecutive terms
3.Fibonacci Sequence
•Each term=sum of previous two terms
Example:0,1,1,2,3,5,8,…
4.Harmonic Progression(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
1.Draw sequences on number lines to visualize progression
2.Check common difference / ratio first before solving
3.For sums, always identify first term (a) and last term (l) or use formula
4.For inserting AMs, divide the gap by n+1
5.For challenging GP problems, use division method to find ratio r.
Visit this link for further practice !!
https://besidedegree.com/exam/s/academic