Question

if a b c are in ap then prove that ab(a+b) bc(b+c) ca(c+a) are in ap​

Answer 1

Answer:

ANSWER

∵a,b,careinAPhence,(b−a)=(c−b)⟶(1)

(i) ∵b2(c+a)−a2(b+c)=(b−a)(ab+bc+ca)andc2(a+b)−b2(c+a)=(c−b)(ab+bc+ca)=(b−a)(ab+bc+ca)(from(1))hence,b2(c+a)−a2(b+c)=c2(a+b)−b2(c+a)∴a2(b+c),b2(c+a),c2(a+b)arealsoinAP

(ii) ∵(c+a−b)−(b+c−a)=2(a−b)=−2(b−a)and(a+b−c)−(c+a−b)=2(b−c)=−2(c−

Answer 2

Sorry to say that but the given condition is impossible it cannot be proved as it does not satisfy the basic criteria for a sequence to be an Arithmetic Progression (AP). You can also check it by replacing the variables a, b, and c by certain numerical values.