Given (5a-3b) /a=(4a+b-2c) /(a+4b-2c) =(a+2b-3c) /(4a-4c)
Let each constant be equal to 'k'
=> (5a-3b) /a = k
=> (5a-3b) = ka-------(1)
(4a+b-2c) /(a+4b-2c) = k
=>(4a+b-2c) = k(a+4b-2c)----(2)
(a+2b-3c) /(4a-4c) = k
=>(a+2b-3c) =k(4a-4c)-------(3)
Now, Adding (1)-(2)+(3),
We get 2a-2b-c = k(4a-4b-2c)
=>2a-2b-c = 2k(2a-2b-c)
=> k = 1/2
Thus, each ratio, k = 1/2.
Now substituting value of , back in equation(1), we get
b/a = 3/2---(4)
Substituting value of b/a = 3/2 and value of k =1/2 in any one of equations
(2) or (3), we get c/a = 2---(5)
From, (4) and (5), we get
6a =4b = 3c.