Let the nos. be a and b; a,b >0.
Then, in the Usual Notations,
A=(a+b)/2, G=sqrt(ab), and, H=(2ab)/(a+b)...............(star).
Given DatarArr A:G=5:4............(1), and, |G-H|=4/5...........(2)
Knowing that, A>=G>=H", we rewrite (2) as, "G-H=4/5.....(2').
(1) rArr (a+b)/(2sqrt(ab))=5/4 rArr 2(a+b)=5sqrt(ab)
rArr 4(a^2+2ab+b^2)-25ab=0", i.e., "4a^2-17ab+4b^2=0
rArr(a-4b)(4a-b)=0 rArr a=4b, or, b=4a
Case : 1 : a=4b
Then, G=sqrt(ab)=2b, and, H=(2*4b*b)/(4b+b)=8/5b
Hence, by (2'), 2b-8/5b=4/5 rArr 2/5b=4/5 rArr b=2
Thus, in this Case, the Nos. are, 8,and, 2.
Case : 2 : b=4a
We simply notice that the roles of a and b have interchanged, so,
we immediately jump to the conclusion that, in this Case, the reqd.
Nos. would be 2, and, 8.
Enjoy Maths.!
