AM-GM Inequality — Applications
Sequences & Series
14
JEE Qs
8%
Hard
75
min
Master the condition for equality in AM-GM, as it's the key to finding exact extremum values in almost all JEE problems.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always ensure all terms involved in the AM-GM inequality are strictly non-negative; otherwise, the inequality does not directly apply.
- 2The condition for equality (all terms being equal) is critical for finding the exact minimum or maximum value of an expression; if equality is not achievable, the AM-GM only provides a bound.
- 3Look for opportunities to create a constant product (when minimizing a sum) or a constant sum (when maximizing a product) by judiciously choosing terms or performing algebraic manipulations (e.g., adding/subtracting constants, using reciprocals).
- 4For expressions involving `x` and `1/x` (or similar inverse relationships), AM-GM is often highly effective for finding minimum/maximum values.
- 5When a direct application is not obvious, consider splitting terms (e.g., `x` into `x/2 + x/2`) or introducing factors to make the product or sum constant and ensure the equality condition can be met.
⚠️ Common Mistakes
- ✕Applying the AM-GM inequality to terms that can be negative, leading to incorrect conclusions.
- ✕Assuming that the minimum or maximum value obtained from AM-GM is always achievable without verifying if the equality condition (all terms are equal) can actually be satisfied by real numbers.
- ✕Incorrectly manipulating algebraic expressions before applying AM-GM, such as altering the product or sum in a way that invalidates the desired outcome or the equality condition.
📝 Practice Questions
See allQ12.The remainder, when 7103 is divided by 23 , is equal to : (1) 6 (2) 17 (3) 9 (4) 14
Q22.Let a1, a2, … , a2024 be an Arithmetic Progression such that a1 + (a5 + a10 + a15 + … + a2020) + a2024 = 2233. Then a1 + a2 + a3 + … + a2024 is equal to _______ 1 2 3 , then α is equal to ________ (3x + t = 5eα ( 85 )
Q1. Let a1, a2, a3, … be a G.P. of increasing positive terms. If a1a5 = 28 and a2 + a4 = 29, then a6 is equal to: (1) 628 (2) 812 (3) 526 (4) 784 = 0. If x(1) = 1, then x ( 12 ) is :
Q15.If ∑nr=1 Tr = (2n−1)(2n+1)(2n+3)(2n+5)64 , then limn→∞∑nr=1 ( Tr1 ) (1) 0 (2) 23 (3) 1 (4) 13
Q13.Suppose that the number of terms in an A.P. is 2k, k ∈N . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27, then k is equal to : (1) 6 (2) 5 (3) 8 (4) 4 y+2
Q1. If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to (1) −1080 (2) −1020 (3) −1200 (4) −120
NCERT Chapters
- Class 11 Maths Ch 9: Sequences and Series