QJEEM14Ma9d41: JEE Main 2014 Mathematics - Class 11 JEEM14Ma9d41 Solution: The variance of first 50 even natural numbers is

Question

Verified Solution • 2025–26 Session moderate

The variance of first 50 even natural numbers is

Accepted Answer

Here is total

50

numbers, so

N=50

Variance

=

{{\sum {{x^2}} } \over {50}} - {\left( {{{\sum x } \over {50}}} \right)^2}

Here

\sum {{x^2}} =

sum of square of first

50

even natural number.

= {2^2} + {4^2} + ..... + {100^2}

= {2^2}\left[ {{1^2} + {2^2} + ....... + {{50}^2}} \right]

= 4\left[ {{{50 \times 51 \times 101} \over 6}} \right]

So,

{{\sum {{x^2}} } \over {50}} = {{4 \times 51 \times 101} \over 6} = 3434

\sum {x = }

sum of first

50

even natural numbers

= 2 + 4 + ...... + 100

= 2\left[ {1 + 2 + .... + 50} \right]

= 2\left[ {{{50 \times 51} \over 2}} \right]

= 50 \times 51

\therefore\,\,\,

{\left( {{{\sum x } \over {50}}} \right)^2} = {\left( {{{50 \times 51} \over {50}}} \right)^2} = 2601

\therefore\,\,\,

Variance

= 3434 - 2601

=833

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