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a 2 +9b 2 +25c 2 =abc((15/a)+(5/b)+(3/c)),then a,b,c are in a.A.P b.G.P c.H.P d. non of these

a2+9b2+25c2 =abc((15/a)+(5/b)+(3/c)),then a,b,c are in
a.A.P  b.G.P  c.H.P  d. non of these
 

Grade:12

2 Answers

rajat Bansal
20 Points
6 years ago
a+9b2+25c2= 15bc+5ac+3ab
a+ 9b2 +25c2 -15bc-5ac-3ab=0
multyply by 2
2a2+18b2+50c2-30bc-10ac-6ab=0
a2-6ab +9b2 +9b2-30bc+25c2+ 25c2-10ac +a2=0
(a-3b)2 +(3b-5c)2 +(5c-a)2=0
square of any no is always positive and sum of three positive no can’t be zero so
a=3b, 
3b=5c
5c=a
so basically
a=3b=5c
hence
b=a/3
c=a/5
so our nos are a,a/3,a/5 
which are in H.P.
 1/a +5/a =2*3/a
 
 
RAJORSHI PAUL
36 Points
6 years ago
Let x=a,y=3b,z=5c
So the expression reduces to
x2 +y2 +z2=xy+yz+zx
So (x-y)2 +(y-z)2 +(z-x)2=0
So x=y=z
So a=3b=5c=r(say)
So, a=r, b=r/3, c=r/5
which shows a,b,c are in HP (Ans.)

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