⇒b−c=a−b ⇒2b=a+c ⇒ a,b, c are in AP, which is true. Now, let 1/(b+c) ,1/(c+a) and 1/(a+b) are in AP
2
c+a
=
1
b+c
+
1
a+b
⇒2(a+b)(b+c)=(a+b+b+c)(c+a) ⇒2(ab+ac+b2+bc)=(ac+a2+2bc+2ab+c2+ca) ⇒2ab+2ac+2b2+2bc=ac+a2+2bc+2ab+ac+c2 ⇒2b2=a2+c2 a,b,c are not in AP, so this is not true. Hence, option (1) is correct.