Question

If a,b,c are in A.P., then prove that :
{a}^{2}(b + c) + {b}^{2}(c + a) + {c}^{2}(a + b) = \frac{2}{9}{(a + b + c)}^{3}a2(b+c)+b2(c+a)+c2(a+b)=92​(a+b+c)3
​

Answer 1

Answer:

Step-by-step explanation:

∵a,b,careinAP

hence,(b−a)=(c−b)⟶(1)

(i) ∵b  

2

(c+a)−a  

2

(b+c)

=(b−a)(ab+bc+ca)

andc  

2

(a+b)−b  

2

(c+a)

=(c−b)(ab+bc+ca)

=(b−a)(ab+bc+ca)(from(1))

hence,b  

2

(c+a)−a  

2

(b+c)=c  

2

(a+b)−b  

2

(c+a)

∴a  

2

(b+c),b  

2

(c+a),c  

2

(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−b)

=−2(b−a)(from(1))

hence(c+a−b)−(b+c−a)=(a+b−c)−(c+a−b)

∴(b+c−a),(c+a−b),(a+b−c)areinAP

(iii) ∵(ca−b  

2

)−(bc−a  

2

)

=(a−b)(a+b+c)

=−(b−a)(a+b+c)

and(ab−c  

2

)−(ca−b  

2

)

=(b−c)(a+b+c)

=−(c−b)(a+b+c)

=−(b−a)(a+b+c)(from(1))

hence,(ca−b  

2

)−(bc−a  

2

)=(ab−c  

2

)−(ca−b  

2

)

∴(bc−a  

2

),(ca−b  

2

),(ab−c  

2

)areinAP