If 1b−c,1c−a,1a−bbe consecutive terms of an AP, then (b−c)2,(c−a)2,(a−b)2 will be in

Now,  we assume (b−c)2,(c−a)2,(a−b)2 are in AP, then we have(c−a)2−(b−c)2=(a−b)2−(c−a)2⇒ (b−a)(2c−a−b)=(c−b)(2a−b−c)    ......(i)Also,if  1b−c,1c−a,1a−b are in,AP, then1c−a−1b−c=1a−b−1c−ab+a−2c(c−a)(b−c)=c+b−2a(a−b)(c−a)⇒ (a−b)(b+a−2c)=(b−c)(c+b−2a)⇒ (b−a)(2c−a−b)=(c−b)(2a−b−c)which is equal to Eq.(i), so our hypothesis is true.