If a2+2bc,b2+2ac,c2+2ab are in A.P. ,show that 1/(b-c),1/(c-a),1/(a-b) are also in A.P.
1/ b-c , 1/ c-a , 1/ a-b are in AP iff
2/ c-a = 1/a-b + 1/ b-c
= b-c + a-b / (a-b)(b-c
2/ c-a = (a-c )/ (a-b )(b-c)
2(a-b)(b-c)= -(c-a)2
2ab - 2ac - 2b2 + 2bc = -c2 - a2 + 2ac
2(b2 + 2ac )= (c2 + 2ab + a2 + 2bc )
which means
which is true