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# CAT 2021 Quant Question [Slot 1] with Solution 18

Question
If ${x_0} = 1,{x_1} = 2$, and ${x_{n + 2}} = \frac{{1 + {x_{n + 1}}}}{{{x_n}}},n = 0,1,2,3,......,$ then ${x_{2021}}$ is equal to
1. 4
2. 1
3. 3
4. 2
Option: 4
Solution:
${x_0} = 1$
${x_1} = 2$
${x_2} = \frac{{\left( {1 + {x_1}} \right)}}{{{x_0}}} = \frac{{\left( {1 + 2} \right)}}{1} = 3$
${x_3} = \frac{{\left( {1 + {x_2}} \right)}}{{{x_1}}} = \frac{{\left( {1 + 3} \right)}}{2} = 2$
${x_4} = \frac{{\left( {1 + {x_3}} \right)}}{{{x_2}}} = \frac{{\left( {1 + 2} \right)}}{3} = 1$
${x_5} = \frac{{\left( {1 + {x_4}} \right)}}{{{x_3}}} = \frac{{\left( {1 + 1} \right)}}{2} = 1$
${x_6} = \frac{{\left( {1 + {x_5}} \right)}}{{{x_4}}} = \frac{{\left( {1 + 1} \right)}}{1} = 2$
Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5n+1 are 2... and so on, where n=0,1,2,3,....
2021 is of the form 5n+1. Hence, its value will be 2.
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