Question : For real $a, b, c$ if $a^2+b^2+c^2=ab+bc+ca$, then value of $\frac{a+c}{b}$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 0
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Correct Answer: 2
Solution :
Given: $a^{2} + b^{2} + c^{2} = (ab+bc+ca)$
⇒ $a^{2} + b^{2} + c^{2} - (ab+bc+ca)=0$
Since : $a^2+b^2+c^2-ab-bc-ca = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]$
So, $a^2+b^2+c^2-ab-bc-ca = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]=0$
⇒ $a=b=c $
So, $\frac{a+c}{b} = \frac{a+a}{a} = 2$
Hence, the correct answer is 2.
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