Question

Anyone please answer this question.​

Answer 1

Given--->

(b+c)/a , (c+a)/b ,(a+b)/c are in AP

Solution-->

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If x , y , z are in AP then if we add same real number to all terms then resulting number are also in Ap

To understand this concept if we add p to all terms then resulting terms are also in AP

It means if x , y , z are in AP then

x+p , y+p , z+p are in AP

Similarly we can divide these terms by same number then resulting number are also in A.P.

It means if x , y , z are in A.P. then

x / p , y / p , z / p are also in A.P.

Now returning to original problem

It is given that

(b+c)/a , (c+a)/b , (a+b)/c are in A.P.

Adding '1' to each term we get

{(b+c)/a + 1} , {(c+a)/b + 1} , {(a+b)/c + 1}

are in A.P.

Taking LCM a , b , c in first , second , third term respectively we get

(b+c+a) / a , (c+a+b)/b , (a+b+c)/c are in A.P.

rearranging the numerators of first two terms we get

(a+b+c)/a , (a+b+c)/b , (a+b+c)/c are in A.P.

Now numerators of all the terms are

(a + b + c)

So dividing all the terms by (a + b + c ) we get.

1 / a , 1 / b , 1 / c are in A.P.