Question

If $a,b$ and $c$ are in A.P. Show that : $(b+c-a),(c+a-b)$ and $(a+b-c)$ are also in A.P.

Answer 1

a, b and c are in AP

So, $2b=a+c...\left(1\right)$

Now,

$x=b+c-a$

$y=c+a-b$

$z=a+b-c$

$2y=2(c+a-b)$

$2y=2(2b-b)$ [from (i)]

$2y=2b$

$x+z=b+c-a+a+b-c$

$x+y=2b$

$2y=x+z$

This proves

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