If (b+c-a)/a , (c+a-b)/b & (a+b-c)/c are in A.P., then prove that 1/a, 1/b & 1/c are also in A.P. ...... answer urgently needed!!

add 2 to each term then divide each term by (a+b+c)

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adding 1 ,b+c-a /a  + 1 ,c+a-b /b +1 , a+b-c/c + 1 =b+c/a ,c+a/b ,a+b/c.
adding 1 again, we get a+b+c/a ,a+b+c/b ,a+b+c/c
cancelling the numerator we get 1/a,1/b,1/c which are thus  also in A.P.
 
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b+c-a/a, c+a-b/b, a+b-c/c in AP 

+%28c%2Ba-b%29%2Fb+-+%28b%2Bc-a%29%2Fa+=+%28a%2Bb-c%29%2Fc+-+%28c%2Ba-b%29%2Fb 

+%28a%28c%2Ba-b%29+-+b%28b%2Bc-a%29%29%2Fab+=+%28b%28a%2Bb-c%29-c%28c%2Ba-b%29%29%2Fbc 

%28ac%2Ba%5E2-ab-b%5E2-bc%2Bab%29%2Fab+=+%28ab%2Bb%5E2-bc-c%5E2-ac%2Bbc%29%2Fbc 


%28a%5E2-b%5E2%2Bac-bc%29%2Fab+=+%28+b%5E2-c%5E2%2Bab-ac%29%2Fbc 

%28%28a%2Bb%29%28a-b%29%2Bc%28a-b%29%29%2Fab=%28%28b%2Bc%29%28b-c%29%2Ba%28b-c%29%29%2Fbc 

%28%28a-b%29%28a%2Bb%2Bc%29%29%2Fab=+%28%28b-c%29%28a%2Bb%2Bc%29%29%2Fbc 

%28a-b%29%2Fa=+%28b-c%29%2Fc 

ac-bc = ab-ac 
/abc 
1/b -1/a = 1/c-1/b 
Hence they are in AP 
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It's depend on you how you want to solve it
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