Question

a2(b+c), b2(c+a), c2(a+b) are in A.P., then either a,b,c are inA.P. or
rn
1. ab+bc+ca=0 2. a+b+c=0
rn
3. a-b-c=0 4. a-b+c=0

Answer 1

a, b, c are in A.P.

⇒b-a=c-b

⇒2b=c+a..........(i)

Now,

a2(b+c), b2(c+a), c2(a+b) are in A.P.

⇒ b2(c+a)-a2(b+c)= c2(a+b)- b2(c+a)

⇒ b2c+ b2a-a2b- a2c= c2a+ c2b- b2c-a b2

⇒ b2c+ b2c+ b2a+a b2-a2b- a2c-c2a- c2b=0

⇒ 2b2c-a2b+2a b2- c2b=ac(c+a)

⇒ ba(2b-a)+bc(2b-c)=ac(c+a)

⇒ bac+bca=ac(c+a) [From(i)]

⇒ 2abc=ac(c+a)

⇒ 2b=c+a

which is true.

Hence, a, b, c are in A.P.
I hope this will help you
if not then comment me

Answer 2

Step-by-step explanation:

mark as brainiest

☺️