Question

Let the mean of n items be $\overline{x}$. If the ${{r}^{th}}$ term is increased by 2r, find the new mean.

Answer 1

Hint: Assume the given n observations as ${{x}_{1}},{{x}_{2}},{{x}_{3}},.....,{{x}_{n}}$. To find the expression for the mean of these observations, take their sum and divide it by n to form a relation. Now, to find the new observations subtract 2 from ${{x}_{1}}$, 4 form ${{x}_{2}}$, 6 from ${{x}_{3}}$ and carry with the same pattern to subtract 2n from ${{x}_{n}}$. Take the sum of these new observations and divide it by n to get the new mean in terms of $\overline{x}$ and n.

Complete step by step answer:
Here we have been provided with n items and their mean equal to $\overline{x}$. We have been asked to calculate the new mean if the ${{r}^{th}}$ term is increased by 2r. Here r will take different integral values according to which the observations will change.
Now, we know that the mean of some given observations is the ratio of sum of all the observations and the number of observations. Let us assume the provided n observations as ${{x}_{1}},{{x}_{2}},{{x}_{3}},.....,{{x}_{n}}$ so we have,
$\Rightarrow \overline{x}=\dfrac{~{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}}}{n}$ …….. (1)
Now, ${{r}^{th}}$ term is increased by 2r that means ${{1}^{st}}$ term is increased by 2, ${{2}^{nd}}$ by 4, ${{3}^{rd}}$ by 6 and if we continue to follow this pattern the ${{n}^{th}}$ term is increased by 2,. Therefore the new observations will be given as $\left( {{x}_{1}}+2 \right),\left( {{x}_{2}}+4 \right),\left( {{x}_{3}}+6 \right),......,\left( {{x}_{n}}+2n \right)$. Assuming the new mean as $\overline{X}$ and taking the ratio of the sum of new observations to the number of observations we get,
\[\begin{align}
  & \Rightarrow \overline{X}=\dfrac{~\left( {{x}_{1}}+2 \right)+\left( {{x}_{2}}+4 \right)+\left( {{x}_{3}}+6 \right)+....+\left( {{x}_{n}}+2n \right)}{n} \\
 & \Rightarrow \overline{X}=\dfrac{~\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}} \right)+\left( 2+4+6+....+2n \right)}{n} \\
\end{align}\]
Clearly we can see that the sum (2 + 4 + 6 + ….. + 2n) is the sum of first n even natural numbers whose simplified sum is given by the formula $n\left( n+1 \right)$, so we get,
\[\begin{align}
  & \Rightarrow \overline{X}=\dfrac{~\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}} \right)+n\left( n+1 \right)}{n} \\
 & \Rightarrow \overline{X}=\dfrac{~\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}} \right)}{n}+\dfrac{n\left( n+1 \right)}{n} \\
\end{align}\]
Using equation (1) and simplifying we get,
\[\therefore \overline{X}=\overline{x}+n+1\]
Hence, the above expression is the new mean of n new observations.

Note: You must remember the formulas of sum of first n even and odd natural numbers which is derived using the concept of arithmetic progression. The sum of first n odd natural numbers is equal to ${{n}^{2}}$. In the progression of first n odd natural numbers the first term is 1 and the common difference is 2 while in the progression of first n even natural numbers the first term is 2 and the common difference is also 2.