Question 148

If $$a+\frac{1}{b}=1$$ and $$b+\frac{1}{c}=1$$, then $$c+\frac{1}{a}$$ is equal to:

Solution

Given, $$a+\frac{1\ }{b}=1$$

$$=$$> $$a=1-\frac{1\ }{b}$$

$$=$$> $$a=\frac{b-1\ }{b}$$

$$=$$> $$\frac{1\ }{a}=\frac{b\ }{b-1}$$

Also, $$b+\frac{\ 1}{c}=1$$

$$=$$> $$\frac{\ 1}{c}=1-b$$

$$=$$> $$c=\frac{\ 1}{1-b}$$

$$\therefore c+\frac{\ 1}{a}=\frac{\ 1}{1-b}+\frac{\ b}{b-1}=\frac{\ 1}{1-b}-\frac{\ b}{1-b}=\frac{\ 1-b}{1-b}=1$$

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