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

10
2 $\frac{1}{2}$ $\frac{1}{2}$
31
42

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + \frac{1}{b} = 1 \\ \Rightarrow a b + 1 = b \\ \Rightarrow a b - b = - 1 \\ \Rightarrow b (a - 1) = - 1 \\ \Rightarrow b = \frac{1}{(1 - a)} . \\ b + \frac{1}{c} = 1 \\ \Rightarrow b c + 1 = c \\ \Rightarrow b c - c = - 1 \\ \Rightarrow c (b - 1) = - 1 \\ \Rightarrow c = \frac{1}{(1 - b)} \\ ∴ c + \frac{1}{a} \\ = \frac{1}{(1 - b)} + \frac{1}{a} \\ = \frac{1}{1 - (\frac{1}{1 - a})} + \frac{1}{a} \\ = \frac{1}{\frac{(1 - a) - 1}{(1 - a)}} + \frac{1}{a} \\ = \frac{(1 - a)}{- a} + \frac{1}{a} \\ = \frac{(a - 1)}{a} + \frac{1}{a} \\ = \frac{a - 1 + 1}{a} \\ = \frac{a}{a} \\ = 1 \end{aligned}

$\eqalign{ & a + \frac{1}{b} = 1{\text{ }} \cr & \Rightarrow ab + 1 = b \cr & \Rightarrow ab - b = - 1 \cr & \Rightarrow b\left( {a - 1} \right) = - 1 \cr & \Rightarrow b = \frac{1}{{\left( {1 - a} \right)}}. \cr & b + \frac{1}{c} = 1 \cr & \Rightarrow bc + 1 = c \cr & \Rightarrow bc - c = - 1 \cr & \Rightarrow c\left( {b - 1} \right) = - 1 \cr & \Rightarrow c = \frac{1}{{\left( {1 - b} \right)}} \cr & \therefore c + \frac{1}{a} \cr & = \frac{1}{{\left( {1 - b} \right)}} + \frac{1}{a} \cr & = \frac{1}{{1 - \left( {\frac{1}{{1 - a}}} \right)}} + \frac{1}{a} \cr & = \frac{1}{{\frac{{\left( {1 - a} \right) - 1}}{{\left( {1 - a} \right)}}}} + \frac{1}{a} \cr & = \frac{{\left( {1 - a} \right)}}{{ - a}} + \frac{1}{a} \cr & = \frac{{\left( {a - 1} \right)}}{a} + \frac{1}{a} \cr & = \frac{{a - 1 + 1}}{a} \cr & = \frac{a}{a} \cr & = 1 \cr}$