Question : If $a+b=2c$, then the value of $\frac{a}{a–c}+\frac{c}{b–c}$ is equal to (where $a\neq b\neq c$):
Option 1: $–1$
Option 2: $1$
Option 3: $0$
Option 4: $\frac{1}{2}$
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Correct Answer: $1$
Solution : Given: $a+b=2c$ ⇒ $a-c=c-b$ Now, $\frac{a}{a-c}+\frac{c}{b-c}$ can be written as, $=\frac{a}{a-c}-\frac{c}{c-b}$ $=\frac{a}{a-c}-\frac{c}{a-c}$ $=\frac{a-c}{a-c}= 1$ Hence, the correct answer is $1$.
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Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}$ where $a \neq b\neq c\neq 0$, then the value of $a^{2}b^{2}c^{2}$ is:
Question : If $a+\frac{1}{a}=1$, then the value of $\frac{a^2-a+1}{a^2+a+1}$ is $(a\neq 0)$:
Question : If $x+\frac{1}{x}=c+\frac{1}{c}$, then the value of $x$ is:
Question : If $\frac{a}{1-2a}+\frac{b}{1-2b}+\frac{c}{1-2c}=\frac{1}{2}$, then the value of $\frac{1}{1-2a}+\frac{1}{1-2b}+\frac{1}{1-2c}$ is:
Question : If $\frac{a^2+b^2}{c^2}=\frac{b^2+c^2}{a^2}=\frac{c^2+a^2}{b^2}=\frac{1}{k}$, $(k\neq 0)$, then $k$=?
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