are in HP. ⇒b+c,c+a,a+b are in AP. ⇒(a+b+c)−(b+c),(a+b+c) −(c+a),(a+b+c)−(a+b) are in AP. ⇒a,b,c are in AP. 2. From 1; a,b,c are in AP. ∴b=a+d,c=a+2d where, d is common difference. ∴(b+c)2=(a+d+a+2d)2 =(2a+3d)2 (c+a)4=(a+2d+a)4=(2a+2d)4 (a+b)2=(a+a+d)2=(2a+d)2 Here, (c+a)4=(b+c)2⋅(a+b)2 So, (b+c)2,(c+a)2,(a+b)2 are not in G.P. Hence, option (a) is correct.