Which relation is true for A ?

If A = $\frac{1}{\sqrt{1} + \sqrt{3}}$ + $\frac{1}{\sqrt{3} + \sqrt{5}}$ + $\frac{1}{\sqrt{5} + \sqrt{7}}$ + ...... upto 50 terms, then which of the following relation is true for A : 4 $\le$ A $\le$ 4.5 9 $\le$ A $\le$ 10 4.5 $\le$ A$\le$ 5 None of the above

Jul 20, 2018

$\text{The right option is } \left(3\right) 4.5 \le A \le 5$.

Explanation:

$A = \frac{1}{\sqrt{3} + \sqrt{1}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \ldots 50 \text{terms}$.

If ${A}_{n}$ is the ${n}^{t h}$ term, then, ${A}_{n} = \frac{1}{\sqrt{2 n + 1} + \sqrt{2 n - 1}}$,

$= \frac{\sqrt{2 n + 1} - \sqrt{2 n - 1}}{\left(2 n + 1\right) - \left(2 n - 1\right)}$.

$\Rightarrow {A}_{n} = \frac{1}{2} \left\{\sqrt{2 n + 1} - \sqrt{2 n - 1}\right\}$.

$\therefore A = {A}_{1} + {A}_{2} + {A}_{3} + \ldots + {A}_{49} + {A}_{50}$,

=1/2{(cancelsqrt3-sqrt1)+(cancelsqrt5-cancelsqrt3)+(cancelsqrt7-cancelsqrt5)+...

+(cancelsqrt99-cancelsqrt97)+(sqrt101-cancelsqrt99)}.

$\therefore 2 A = \sqrt{101} - 1. \ldots \ldots \ldots . . \left(\ast\right)$.

Now, $100 < 101 < 121$.

$\therefore \sqrt{100} < \sqrt{101} < \sqrt{121} ,$.

$\mathmr{and} , 10 < \sqrt{101} < 11$.

$\text{Adding" -1," we get, } 9 < \sqrt{101} - 1 < 10$.

$\therefore \left(\ast\right) \Rightarrow 9 < 2 A < 10$.

$\text{Dividing by "2 gt 0," we get, } 4.5 < A < 5$.

Hence, the right option is $\left(3\right) 4.5 \le A \le 5$.