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HP — nth term, HM, AM-GM-HM inequality

Sequences & Series

14

JEE Qs

8%

Hard

75

min

Master the transformation of HP to AP, understand the conditions for AM-GM-HM inequality, and practice its application for finding min/max values.

🧮 Key Formulas

If a_1, a_2, ..., a_n are in HP, then 1/a_1, 1/a_2, ..., 1/a_n are in AP.
nth term of HP: T_n = 1 / (A_1 + (n-1)D), where A_1 is the first term of the corresponding AP and D is its common difference.
Harmonic Mean (HM) of two numbers a, b: HM = 2ab / (a+b)
Harmonic Mean (HM) of n numbers x_1, x_2, ..., x_n: HM = n / (1/x_1 + 1/x_2 + ... + 1/x_n)
AM-GM-HM Inequality (for positive numbers x_1, x_2, ..., x_n): (x_1 + ... + x_n)/n >= (x_1 * ... * x_n)^(1/n) >= n / (1/x_1 + ... + 1/x_n)
For two positive numbers a, b: AM * HM = GM^2

✅ Key Points for JEE

  • 1The most effective way to solve problems involving HP is to convert them into corresponding AP problems by taking the reciprocal of each term.
  • 2The AM-GM-HM inequality is strictly applicable only for *positive* real numbers. Always check this condition before application.
  • 3Equality in the AM-GM-HM relation (AM = GM = HM) holds true if and only if all the numbers involved are equal.
  • 4AM-GM inequality is a powerful tool for finding the minimum value of a sum (when product is constant) or maximum value of a product (when sum is constant) in optimization problems.
  • 5Harmonic Mean often appears in problems involving rates (e.g., average speed, average work rate) where the variable is in the denominator.

⚠️ Common Mistakes

  • ✕Attempting to apply AP or GP formulas directly to an HP series without converting it to its reciprocal AP form.
  • ✕Forgetting the crucial condition that numbers must be *positive* when applying AM-GM-HM inequality, leading to incorrect results.
  • ✕Incorrectly identifying when equality holds (all terms equal) in AM-GM-HM inequality, which is vital for finding exact min/max values.
  • ✕Algebraic errors in manipulating reciprocals or combining terms after applying AM-GM-HM.

📝 Practice Questions

See all

Q12.The remainder, when 7103 is divided by 23 , is equal to : (1) 6 (2) 17 (3) 9 (4) 14

2025·MCQMedium

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 )

2025·NumericalMedium

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 :

2025·MCQMedium

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

2025·MCQHard

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

2025·MCQMedium

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

2025·MCQEasy

NCERT Chapters

  • Class 11 Maths Ch 9: Sequences and Series