Question

Plse solve this sum

http://www.goiit.com/posts/list/about-iits-and-jee-if-the-a-m-g-m-h-m-of-first-last-terms-of-1018425.htm

Answer 1

first and last terms of \(25,26,...,N-1,N\) are : \(25,N\)\[A.M: \frac{25+N}{2}\]\[G.M: \sqrt{25N}\]\[H.M: \frac{25N}{25+N}\]right?

Answer 2

sorry\[H.M: \frac{50N}{25+N}\]

Answer 3

and do u know that \(A.M\ge G.M \ge H.M\) ?? :)

Answer 4

yes

Answer 5

very well, so how about this:\[ \frac{25+N}{2}-2=\sqrt{25N}-1=\frac{50N}{25+N}\]

Answer 6

now, we have to compare these terms so that we can get the value of N

Answer 7

yes id probably say first compare first with third

Answer 8

okay

Answer 9

factors are not coming

Answer 10

how you subtract 2 from (25+N/2)

Answer 11

sorry i was out :) thats what question says

Answer 12

always before solving problems, read the question carefully :)

Answer 13

okay , i got it

Answer 14

i hope that helps :)

Answer 15

yes it helps
but value of N is not coming

Answer 16

lets check it\[\frac{25+N}{2}-2=\frac{50N}{25+N}\]\[2(25+N)^2-4(25+N)=100N\]u r right no whole number for N from this equation

Answer 17

so what to do

Answer 18

because it makes an equation
N^2-54N+525=0

Answer 19

do u have the answers?

Answer 20

no
i found this question on internet

Answer 21

because we did it right i think :\

Answer 22

which i have pasted

Answer 23

i think we should wait for others opinions...

Answer 24

okay