Question

find mean deviation about mean for first 2n + 1 natural numbers​

Answer 1

$\displaystyle\large\boxed{\sf{MD(\bar x)=\dfrac{n(n+1)}{2n+1}}}$

The formula for mean deviation about mean is,

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{n}\sum_{i=1}^n|x_i-\bar x|}$

where $\displaystyle\sf{x_i}$ is the $\displaystyle\sf{i^{th}}$ observation and $\displaystyle\sf{\bar x}$ is the mean.

But here,

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2n+1}\sum_{i=1}^{2n+1}|x_i-\bar x|}$

The mean of first $\displaystyle\sf{2n+1}$ natural numbers, if they are considered as in an AP, is,

$\displaystyle\longrightarrow\sf{\bar x=\dfrac{\dfrac{2n+1}{2}\Big[1+2n+1\Big]}{2n+1}}$

$\displaystyle\longrightarrow\sf{x=\dfrac{2n+2}{2}}$

$\displaystyle\longrightarrow\sf{\bar x=n+1}$

The mean can also be found by standard formula for mean, i.e.,

$\displaystyle\longrightarrow\sf{\bar x=\dfrac{1}{2n+1}\sum_{i=1}^{2n+1}i}$

$\displaystyle\longrightarrow\sf{\bar x=\dfrac{1}{2n+1}\cdot\dfrac{(2n+1)(2n+2)}{2}}$

$\displaystyle\longrightarrow\sf{\bar x=n+1}$

Well, it is the middle term, i.e., the median.

It is true that the mean deviation of two particular terms having similar positions from either side of the sequence is the same (but in the case of sequences containing consecutive terms). For example, here, since there are $\displaystyle\sf{2n+1}$ terms,

• Mean deviation of $\displaystyle\sf{1^{st}}$ term about mean is the same as that of $\displaystyle\sf{(2n+1)^{th}}$ term.

• Mean deviation of $\displaystyle\sf{2^{nd}}$ term about mean is the same as that of $\displaystyle\sf{(2n)^{th}}$ term.

• Mean deviation of $\displaystyle\sf{3^{rd}}$ term about mean is the same as that of $\displaystyle\sf{(2n-1)^{th}}$ term.

Generally,

• Mean deviation of $\displaystyle\sf{i^{th}}$ term about mean for any natural number $\displaystyle\sf{1\leq i\leq2n+1}$ is the same as that of $\displaystyle\sf{(2n+2-i)^{th}}$ term.

Therefore,

$\displaystyle\longrightarrow\sf{\sum_{i=1}^{n+1}|x_i-\bar x|=\sum_{i=n+1}^{2n+1}|x_i-\bar x|}$

Here, $\displaystyle\sf{x_i=i}$ and $\displaystyle\sf{\bar x=n+1.}$ Then,

$\displaystyle\longrightarrow\sf{\sum_{i=1}^{n+1}|i-(n+1)|=\sum_{i=n+1}^{2n+1}|i-(n+1)|\quad\quad\dots(1)}$

Now, the mean deviation about mean,

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2n+1}\sum_{i=1}^{2n+1}|x_i-\bar x|}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2n+1}\sum_{i=1}^{2n+1}|i-(n+1)|}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2n+1}\left[\sum_{i=1}^{n+1}|i-(n+1)|+\sum_{i=n+1}^{2n+1}|i-(n+1)|\right]}$

From (1),

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{2}{2n+1}\sum_{i=1}^{n+1}|i-(n+1)|\quad\quad\dots(2)}$

But for $\displaystyle\sf{1\leq i\leq(n+1),\quad i-(n+1)\leq0.}$ Then (2) becomes,

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{2}{2n+1}\sum_{i=1}^{n+1}(n+1-i)}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{2}{2n+1}\left[\sum_{i=1}^{n+1}(n+1)-\sum_{i=1}^{n+1}i\right]}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{2}{2n+1}\left[(n+1)^2-\dfrac{(n+1)(n+2)}{2}\right]}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{2(n+1)}{2n+1}\left[n+1-\dfrac{n+2}{2}\right]}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{2(n+1)}{2n+1}\cdot\dfrac{(2n+2)-(n+2)}{2}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{MD(\bar x)=\dfrac{n(n+1)}{2n+1}}}}$

Well, let me further go on!

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2}\cdot\dfrac{2n(n+1)}{n+n+1}}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2}\cdot\dfrac{2}{\left(\dfrac{n+n+1}{n(n+1)}\right)}}$

$\displaystyle\longrightarrow\sf{MD(\bar x)=\dfrac{1}{2}\cdot\dfrac{2}{\left(\dfrac{1}{n}+\dfrac{1}{n+1}\right)}}$

The term $\displaystyle\sf{\dfrac{2}{\left(\dfrac{1}{n}+\dfrac{1}{n+1}\right)}}$ is the harmonic mean of $\displaystyle\sf{n}$ and $\displaystyle\sf{n+1.}$

$\displaystyle\longrightarrow\sf{HM(a,\ b)=\dfrac{2}{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}=\dfrac{2ab}{a+b}}$

Therefore,

$\displaystyle\longrightarrow\sf{\underline{\underline{MD(\bar x)=\dfrac{HM(n,\ n+1)}{2}}}}$

Thus we obtain a statement that,

$\boxed{\begin{minipage}{10cm}\sf{The mean deviation of first some odd no. of consecutive natural numbers about the mean is half the harmonic mean of their median and its predecessor.}\\\\\begin{center} \displaystyle\mathsf{MD_{2n+1}(\bar x)=\dfrac{HM(n,\ n+1)}{2}} \end{center}\end{minipage}}$

Answer 2

Step-by-step explanation:

Mean of first n natural no. where n is as

even no is =

2

1

[

2

n

+

2

n+1

]=

4

2n+1

mean deviation=

4n

1

[1+3+5+9...….n

th

odd]

sum of odd no=n

2

=

4n

1

n

2

=

4

n

.

HOPE IT HELPS..