If ratio of AM and GM of two numbers is #5/4#, what are the numbers and what is their HM?

1 Answer
Oct 6, 2017

Answer:

Numbers are #2# and #8#. Their H.M. is #16/5#.

Explanation:

Let the two numbers be #a# and #b#.

Their A.M. is #(a+b)/2# and as G.M. is #sqrt(ab)#

and as ratio between A.M. and G.M. is #5/4#

#((a+b)/2)/(sqrt(ab))=5/4#

or #2a+2b=5sqrt(ab)#

or #4a^2+4b^2+8ab=25ab#

i.e. #4a^2+4b^2-17ab=0#

or #(4a-b)(a-4b)=0#

i.e. either #a/b=4# or #b/a=4#

Hence numbers are of the form #x# and #4x#

and their G.M. is #2x# and H.M.@ will be #(2*x*4x)/(x+4x)# or #(8x^2)/5x# or #(8x/5)#
Further, as difference of G.M. and H.M. is #4/5#

#2x-8/5x=4/5#

or #2/5x=4/5# and #x=2#

Hence numbers are #2# and #8#.

Observe their A.M. is #5#, G.M. is #4# and H.M. is #16/5#.

@ If #h# is H.M. between #a# and #b#,

#1/a,1/h,1/b# are in arithmetic sequence and hence

#2(1/h)=1/a+1/b=(a+b)/(ab)#

and #1/h=(a+b)/(2ab)# i.e. #h=(2ab)/(a+b)#