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If (b+c-a)/a,(c+a-b)/b,(a+b-c)/c are in...

If `(b+c-a)/a,(c+a-b)/b,(a+b-c)/c` are in A.P. Prove that `1/a,1/b,1/c` are in A.P.

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`(b+c - a)/a, (c+a-b)/b and (a+b-c)/c` are in `A.P.`
Now, if we add `2` to all these terms, then again these terms will be in `A.P.`
`:. (b+c - a)/a+2, (c+a-b)/b+2 and (a+b-c)/c+2` are in `A.P.`
`=> (b+c - a+2a)/a, (c+a-b+2b)/b and (a+b-c+2c)/c` are in `A.P.`
`=> (b+c +a)/a, (c+a+b)/b and (a+b+c)/c` are in `A.P.`
If we take, `(a+b+c)` common in these three terms,
Then, `1/a,1/b and 1/c` are in `A.P.`
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