The mean of n items is \pi. If the first term is increased by 1, second by 2 and so on, then new mean is 

  • Option 1)

    \pi +n

  • Option 2)

    \pi +\frac{n}{2}

  • Option 3)

    \pi +\frac{n+1}{2}

  • Option 4)

    None of these 


Answers (1)

As we discussed in concept


For the values x1, x2, ....xn of the variant x the arithmetic mean is given by 

\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}

in case of discrete data.



 Let no are x_{1}, x_{2}---x_{n}

\therefore\:\pi n\:=\:x_{1}+x_{2}+x_{3}+----x_{n}

Now she sum is

= (x_{4}+x_{2}+x_{3}+------+x_{n})+(1+2+3+-----+n)

= (x_{1}+x_{2}+x_{3}+-----x_{n})+\frac{n(n+1)}{2}

=> n\pi +\frac{n(n+1)}{2}

New mean = \frac{1}{n}[n\pi + \frac{n(n+1)}{2}]

                   = \pi + \frac{1}{2}(n+1)

Option 1)

\pi +n

This option is incorrect.

Option 2)

\pi +\frac{n}{2}

This option is incorrect.

Option 3)

\pi +\frac{n+1}{2}

This option is correct.

Option 4)

None of these 

This option is incorrect.

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