Question

if (b+c-a)/a, (c+a-b)/b, (a+b-c) /c are in A.P then show that 1/a,1/b,1/c are in AP
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Answer 1

Step-by-step explanation:

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

Add 2 to all terms

=> (b+c-a)/a + 2, (c+a-b)/b + 2, (a+b-c)/c + 2 are in AP.

=> (b+c-a+2a)/a, (c+a-b+2b)/b , (a+b-c+2c)/c are in AP.

=> (a+b+c)/a, (a+b+c)/b, (a+b+c)/c are in AP.

Divide all terms by (a+b+c)

=> 1/a, 1/b, 1/c are in AP.

=> a, b, c are in HP.

I hope this will help you :)