Question

Q.4

Answer 1

`a,b anc c are in A.P`

`⇒ 2b=a+c`

`"We have to prove that (b+c-a),(c+a-b) and (a+b-c) are A.P"`

That means, we have to prove 

`(b+c+a)+(a+b-c)=2(c+a-b)`

Consider, 

`(b+c-a)+(a+b-c)=b+c-a+a+b-c`

                          ` =2b`

                           `=a+c`

And, `2(c+a-b)=2c+2a-2b`

               `=2c+2a-(a+c)`

                 `=a+c` 

`⇒ (b+c-a)+(a+b-c)=2(c+a-b)`

`⇒(b+c-a),(c+a-b) "and (a+b-c) are also in A.P."`