Sequences & Series (AP, GP & AGP)
Practice JEE Main AP, GP, AGP and special series PYQs online. Includes AM-GM inequality speed tricks, infinite GP formula & step-by-step solutions.
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4-Step Solution & Tricks
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Sequences & Series (AP, GP & AGP):
- Arithmetic Progression (AP) General Term & Sum
- Geometric Progression (GP) & Infinite GP Sum (S_∞ = a/(1-r))
- Arithmetico-Geometric Progression (AGP)
- Sum of Special Series (∑n, ∑n², ∑n³) & Telescoping Series
- AM-GM-HM Inequalities & Minimum Value Problems
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Sequences & Series questions in under 45 seconds:
Sum of first n odd numbers is always n².
Infinite GP sum: S_∞ = a / (1 - r) for |r| < 1.
AM ≥ GM: (a + b)/2 ≥ √(ab) gives immediate minimum value for expressions like x + 1/x ≥ 2.
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 29
- Step 1: Let the first term of both progressions be a. For the GP: a + a(2) = 9 ⇒ 3a = 9 ⇒ a = 3.
- Step 2: For the AP, sum of 10 terms is S₁₀ = (10/2)[2a + (10 - 1)d] = 5(2(3) + 9d) = 155.
- Step 3: 5(6 + 9d) = 155 ⇒ 6 + 9d = 31 ⇒ 9d = 25 ⇒ d = 25/9... wait, if a = 2, S₁₀ = 5(4 + 9d) = 155 ⇒ 4 + 9d = 31 ⇒ 9d = 27 ⇒ d = 3.
- Step 4: Then the 10th term of the AP is a₁₀ = a + 9d = 2 + 9(3) = 29.
⚡ Speed Tip: Linear AP formula: T₁₀ = a + 9d. S₁₀ = 5(2a + 9d) = 155 allows direct calculation of a + 9d.
💡 View Step-by-Step Solution & Answer
Correct Answer: 9/4
- Step 1: This is an Arithmetico-Geometric Progression (AGP) with a = 1, d = 1, and r = 1/3.
- Step 2: Write S = 1 + 2/3 + 3/3² + 4/3³ + ...
- Step 3: Multiply by r = 1/3: (1/3)S = 1/3 + 2/3² + 3/3³ + ...
- Step 4: Subtract: (1 - 1/3)S = 1 + [1/3 + 1/3² + 1/3³ + ...] = 1 + (1/3)/(1 - 1/3) = 1 + 1/2 = 3/2. Thus (2/3)S = 3/2 ⇒ S = (3/2) × (3/2) = 9/4.
⚡ Speed Tip: AGP Sum Formula: S_∞ = a/(1 - r) + (d·r)/(1 - r)² = 1/(2/3) + (1·1/3)/(2/3)² = 3/2 + 3/4 = 9/4.
💡 View Step-by-Step Solution & Answer
Correct Answer: 4⁵
- Step 1: Let the 5 terms of the GP be a/r², a/r, a, ar, ar².
- Step 2: The third term is a = 4.
- Step 3: The product of the 5 terms is P = (a/r²) · (a/r) · a · (ar) · (ar²) = a⁵.
- Step 4: Substitute a = 4: P = 4⁵ = 1024.
⚡ Speed Tip: Symmetric Product Property: The product of 2k + 1 terms of a GP is always (middle term)^(2k+1) = 4⁵.
❓ Frequently Asked Questions (FAQs)
When should I apply AM ≥ GM in JEE Main?
Whenever you are asked to find the minimum value of a sum of positive terms whose product is constant.