If the A.M., the G.M. and the H.M. of the first and the last terms of the series 100, 101, 102,…
…n – 1, n are the terms of the series itself then the value of n is ………(100 < n ≤ 500).
Text Solution
Verified by Experts0400
Ans. 0400
Sol. A.M. = 
, G.M. = 10 
and H.M. = 
For A.M. to be an integer n must be even
For G.M. to be an integer n must be perfect square
⇒ n = 4k 2
∴ H.M. = 
Since numerator is multiple of 25, denominator must be a multiple of 25 as H.M. is an integer.
Also 100 < n ≤ 500
⇒ 25 < k 2 ≤ 125
⇒ k 2 = 50, 75, 100 and 125
Satisfying these values we get k 2 = 100 as the only solution.
⇒ n = 400.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems