Question : If $a^{2}+b^{2}+c^{2}=ab+bc+ca,$ then the value of $\frac{a+c}{b}$ is:
Option 1: 3
Option 2: 2
Option 3: 0
Option 4: 1
Correct Answer: 2
Solution : Given: $a^{2}+b^{2}+c^{2}=ab+bc+ca$ ⇒ $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$ ⇒ $\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0$ ⇒ $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$ The square of any real number is always non-negative. Therefore, we have: $(a-b)^{2}=0 \Rightarrow a=b$ $(b-c)^{2}=0 \Rightarrow b=c$ $(c-a)^{2}=0 \Rightarrow c=a$ Thus, we have $a=b=c$. Now, we can calculate $\frac{a+c}{b}$ as follows: $\frac{a+c}{b}=\frac{a+a}{a}=2$ Hence, the correct answer is 2.
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Question : If $a+b+c=0$, then the value of $\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}$ is:
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