| Statement-1: For every natural number \[\ge 2,\,\,\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+.....+\frac{1}{\sqrt{n}}>\sqrt{n}\]. |
| Statement-2: For every natural number \[n\ge 2,\sqrt{n\left( n+1 \right)}<n+1\]. |
A) Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is false.
C) Statement-1 is false, Statement-2 is true.
D) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Correct Answer: D
Solution :
Statement-2: \[\sqrt{n}<\sqrt{n+1}\] is true for \[n\ge 2\]. Statement-1: \[\sqrt{n}<\sqrt{n+1}\Rightarrow \sqrt{2}<\sqrt{3}<\sqrt{4}<.....\sqrt{n}\] Now \[\sqrt{2}<\sqrt{n}\Rightarrow \frac{1}{\sqrt{2}}>\frac{1}{\sqrt{n}}\] \[\sqrt{3}<\sqrt{n}\Rightarrow \frac{1}{\sqrt{3}}>\frac{1}{\sqrt{n}}\] : : \[\sqrt{n}<\sqrt{n}\Rightarrow \frac{1}{\sqrt{n}}\ge \frac{1}{\sqrt{n}}\] Also \[\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{n}}\]. So\[\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>\frac{n}{\sqrt{n}}=\sqrt{n}\]
You need to login to perform this action.
You will be redirected in
3 sec
