Question

If a, b, c are in A.P., prove that a²(b + c), b²(c + a), c²(a + b) are in A.P. (ab + bc + ca ≠ 0)​

Answer 1

Answer:

0

Step-by-step explanation:

f x,y,z are in A.P. , xy=z−y

a  

2

(b+c)−a  

2

(b+c)=c  

2

(a+b)−b  

2

(c+a)

a  

2

b+a  

2

c−a  

2

b−a  

2

c=c  

2

a+c  

2

b−b  

2

c−b  

2

a

(a  

2

b−a  

2

c)+(b  

2

a−a  

2

b)=(c  

2

a−b  

2

a)+(c  

2

b−b  

2

c)

c(b  

2

−a  

2

)+ab(b−a)=a(c  

2

−b  

2

)+bc(c−b)

(b−a)[c(b+a)+ab]=(c−b)[a(c+b)+bc]

(b−a)(bc+ac+ab)=(c−b)(ac+bc+ab)

Either ab+bc+ac=0,

b−a=c−b

∴a,b,c are in A.P.