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 Data#rArr 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.!