# 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

As we discussed in concept

ARITHMETIC Mean -

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

$\dpi{100} \bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}$

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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