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Given, 5a–3b / a = 4a+b–2c / a+4b–2c = a+2b–3c / 4a–4c. Prove that 6a=4b=3c

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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.

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