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Previous Year Question (PYQs)



The value of $3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$





Solution

Let $x = 4 + \frac{1}{3 + \frac{1}{4 + \cdots}}$ 

Now, $x = 4 + \frac{1}{3 + \frac{1}{x}}$ 

$= 4 + \frac{1}{\frac{3x+1}{x}} = 4 + \frac{x}{3x+1}$ 

$\Rightarrow x(3x+1) = 4(3x+1) + x$ 

$\Rightarrow 3x^2 + x = 12x + 4 + x$ 
$\Rightarrow 3x^2 - 12x - 4 = 0$ 

$\Rightarrow x = \frac{12 + \sqrt{144 + 48}}{6} = \frac{12 + 8\sqrt{3}}{6}$ 

$\Rightarrow x = 2 + \frac{4\sqrt{3}}{3}$ 

Now, $3 + \frac{1}{x} = 3 + \frac{1}{2 + \frac{4\sqrt{3}}{3}}$ 

$= \frac{3}{2} + \sqrt{3}$ 

Final Answer: $(a) 1.5 + \sqrt{3}$


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