Question

If 1/q+r ,1/r+p and 1/p+q are in A.P then show that p square ,q square ,r square are in A.P

Answer 1

1/(q+r) ,1/(r+p) and 1/(p+q) are in AP

hence common difference is constant of this.
=>1/(r+p) - 1/(q+r) = 1/(p+q) -1/(r+p)

=>{q+r-r-p}/(r+p)(q+r) ={r+p-p-q} /(p+q)(r+p)

=>(q-p)/(q+r)=(r-q)/(p+q)

=>(q-p)(p+q)=(r-q)(q+r)

=> q^2-p^2=r^2-q^2

here you see difference between two consecutive term is same
hence ,
p^2 , q^2 , r^2 are in AP