If a,b,care non zero real numbers such that 3(a2+b2+c2+1)=2(a+b+c+ab+bc+ca), thena,b,c are in
Finding the relation of terms a,b,c:
Given a,b,c are non zero real numbers and
3(a2+b2+c2+1)=2(a+b+c+ab+bc+ca)
⇒ 3a2+3b2+3c2+3-2a-2b-2c-2ab-2bc-2ac=0
⇒(a2-2a)+(b2-2b)+(c2-2c)+(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)+1+1+1=0
⇒(a2-2a+1)+(b2-2b+1)+(c2-2c+1)+(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)=0
⇒ (a-1)2+(b-1)2+(c-1)2+(a-b)2+(b-c)2+(c-a)2=0
⇒ (a-1)=0,(b-1)=0,(c-1)=0
⇒ a=1,b=1,c=1
Here 2b=a+c
So condition for A.P is satisfied and
b2=ac, So condition for G.P also satisfied and
2b=(a+c)ac So condition for H.P also satisfied
Therefore the terms a,b,c are in A.P,G.P,H.P
Hence the terms a,b,c are in A.P,G.P,H.P.
If a, b, c are rational numbers such that a2 + b2 + c2 − ab − bc − ca = 0, prove that a = b = c.