Progression (Sequence and Series)HardBloom L3

Question

Let $S = \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \frac{1}{\sqrt{9} + \sqrt{11}} + \ldots\frac{1}{\sqrt{9997} + \sqrt{9999}}$, then

Options

A.$S < 24$
B.$S > 24$
C.$S > 18$
D.$S < 18$

Solution

Sol. [Key Concept: Evaluate the defined sequence/sum.]

$$S = \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \ldots + \frac{1}{\sqrt{9997} + \sqrt{9999}}$$

$$\begin{matrix} & \ > \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{7} + \sqrt{9}} + \ldots\frac{1}{\sqrt{9999} + \sqrt{10001}} \\ \Rightarrow \ 2\text{ }S & \ > \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \ldots + \frac{1}{\sqrt{9999} + \sqrt{10001}} \\ & \ = \frac{1}{2}(\sqrt{3} - \sqrt{1} + \sqrt{5} - \sqrt{3} + \sqrt{7} - \sqrt{5} + \ldots + \sqrt{10001} - \sqrt{9999}) \\ & \ = \frac{1}{2}(\sqrt{10001} - \sqrt{1}) > \frac{1}{2}(100 - 1) \\ \Rightarrow \ S & \ > \frac{99}{4} > 24 \end{matrix}$$

✘ (a) $S > 24$

✔ (b) $2S > \frac{12(}{\sqrt{10001}} - 1) \Rightarrow \ S > 24$

✔ (c) $S > 24 > 18$

✘ (d) $S > 18$

Create a free account to view solution

View Solution Free
Topic: Progression (Sequence and Series)·Practice all Progression (Sequence and Series) questions

More Progression (Sequence and Series) Questions