MCQ
The $S.D$. of the first $n$ natural numbers is
- A$\frac{{n + 1}}{2}$
- B$\sqrt {\frac{{n(n + 1)}}{2}} $
- ✓$\sqrt {\frac{{{n^2} - 1}}{{12}}} $
- DNone of these
$ = \sqrt {\frac{{n(n + 1)(2n + 1)}}{{6n}} - {{\left[ {\frac{{n(n + 1)}}{{2n}}} \right]}^2}} $
$ = \sqrt {\frac{{(n + 1)(2n + 1)}}{6} - {{\left( {\frac{{n + 1}}{2}} \right)}^2}} = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{2n + 1}}{3} - \frac{{n + 1}}{2}} \right)} $
$ = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{4n + 2 - 3n - 3}}{6}} \right)} $
$ = \sqrt {\frac{{{n^2} - 1}}{{12}}} $.
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