Question

If the first item is increased by $1$, second by $2$ and so on, then the new mean is

(Given : Initial mean of all items is $\overline{X}$ and there are $n$ items)

• A. $\overline{X}+n$
• B. $\overline{X}+\frac{n}{2}$
• C. $\overline{X}+\frac{n+1}{2}$
• D. None of these

Answer 1

The correct option is C $\overline{X}+\frac{n+1}{2}$

Given that mean of $n$ numbers is $\overline{X}$

Let the numbers be $a_1,a_2,a_3,...,a_{n-1},a_n$

Therefore the mean $=\frac{a_1+a_2+a_3+...+a_{n-1}+a_n}{n}=\overline{X}$

$⟹a_1+a_2+a_3+...+a_{n-1}+a_n=n\overline{X}$ ————(1)

Given that first term is increased by $1=a_1+1$

Second term is increased by $2=a_2+2$

Similarly $n^{th}$ term is increased by $n=a_n+n$

Therefore new mean $=\frac{a_1+1+a_2+2+a_3+3+...+a_n+n}{n}$ ———-(2)

Substituting (1) in (2) we get

$\text{new mean}=\frac{n\overline{X}+1+2+3+...+n}{n}$

We know that sum of first $n$ natural numbers is $\frac{n(n+1)}{2}$

$⟹\text{new mean}=\frac{n\overline{X}+\frac{n(n+1)}{2}}{n}$

$⟹\text{new mean}=\frac{n\overline{X}}{n}+\frac{n(n+1)}{2n}$

$⟹\text{new mean}=\overline{X}+\frac{n+1}{2}$