Correct Answer - Option 3 : 50

__Concept:__

Suppose there are ‘n’ observations {\({\rm{\;}}{{\rm{x}}_{\rm{1}}},{{\rm{x}}_{{\rm{2\;}}}},{{\rm{x}}_{{\rm{3\;}}}}, \ldots ,{{\rm{x}}_{{\rm{n\;}}}}\)}

Mean \(\left( {{\rm{\bar x}}} \right) = \frac{{{\rm{\;}}({{\rm{x}}_1} + {{\rm{x}}_{2{\rm{\;}}}} + {{\rm{x}}_{3{\rm{\;}}}} + \ldots + {{\rm{x}}_{{\rm{n\;}}}})}}{{\rm{n}}}\) \( = {\rm{\;}}\frac{{\mathop \sum \nolimits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Sum of the first n natural numbers = \(\rm \frac{n(n+1)}{2}\)

__Calculation:__

To find: Mean of the first 99 natural numbers

As we know, Sum of first n natural numbers = \(\rm \frac{n(n+1)}{2}\)

Now, Mean = \(\rm \dfrac { \frac{99(99+1)}{2}}{99}\)

= \(\rm \frac{(99+1)}{2}=50\)