If a ,b, c are in ap a2 (b+c) , b2 (a+c) , c2 (a+b)
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Answer:
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
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