Find out two numbers whose arithemetic mean exeeds their geometric mean by 2 and whose harmonic mean is one fifth of the larger number
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Let the greater number is a and smaller number is b.
AM=GM+2
a+b/2=root (ab)+2
a+b=2(root(ab)+2)
a+b=2 root(ab)+4 ....(1)
HM = a/5
2ab/a+b = a/5
2b*5 = a+b
10b = a+b
10b - b = a
a = 9b .....(20
substitute (2) in (1)
9b+b = 2root (ab) + 4
10b = 2 root (9b*b)+4
10b=2 root(9b^2)+4
10b = 2*3b + 4
10b - 6b = 4
4b = 4
b = 1
substitute in (2) we get
a = 9b
a = 9*1
a = 9
therefore the numbers are a = 9;b = 1
AM=GM+2
a+b/2=root (ab)+2
a+b=2(root(ab)+2)
a+b=2 root(ab)+4 ....(1)
HM = a/5
2ab/a+b = a/5
2b*5 = a+b
10b = a+b
10b - b = a
a = 9b .....(20
substitute (2) in (1)
9b+b = 2root (ab) + 4
10b = 2 root (9b*b)+4
10b=2 root(9b^2)+4
10b = 2*3b + 4
10b - 6b = 4
4b = 4
b = 1
substitute in (2) we get
a = 9b
a = 9*1
a = 9
therefore the numbers are a = 9;b = 1
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