Consider first natural number when $n$ is an odd i.e., 1, 2, 3, 4, ...n, [odd].

$$\begin{aligned} & \text { Mean } \bar{x}= \frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ \therefore \quad & \mathrm{MD}= \left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|+\ldots+\left|n-\frac{n+1}{2}\right| \end{aligned}$$

$$=\frac{\left|-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\ldots+\left|\frac{n-1}{2}-\frac{n+1}{2}\right|+\left|\frac{n+1}{2}-\frac{n+1}{2}\right|+\left|\frac{n+3}{2}-\frac{n+1}{2}\right|+\ldots+\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|n-\frac{n+1}{2}\right|}{n}$$

$$\begin{aligned} & =\frac{2}{n}\left[1+2+\ldots+\frac{n-3}{2}+\frac{n-1}{2}\right] \quad\left(\frac{n-1}{2}\right) \text { terms } \\ & =\frac{2}{n}\left[\frac{\left(\frac{n-1}{2}\right)\left(\frac{n-1}{2}+1\right)}{2}\right] \quad\left[\because \text { sum of first } n \text { natural numbers }=\frac{n(n+1)}{2}\right] \\ & =\frac{2}{n} \cdot \frac{1}{2}\left[\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\right]=\frac{1}{n}\left(\frac{n^2-1}{4}\right)=\frac{n^2-1}{4 n} \end{aligned}$$

Consider first $n$ natural number, when $n$ is even i.e., $1,2,3,4,\quad$ [even]

$$\begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ & \mathrm{MD}=\frac{1}{n}\left[\left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|\right]+\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|\frac{n}{2}-\frac{n+1}{2}\right| \\ &+\left|\frac{n+2}{2}-\frac{n+1}{2}\right|+\ldots+\left|n-\frac{n+1}{2}\right| \\ &=\frac{1}{n}\left[\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+\ldots .+\left|\frac{-3}{2}\right|+\left|\frac{1}{2}\right|+\ldots+\left|\frac{n-1}{2}\right|\right] \\ &=\frac{2}{n}\left[\frac{1}{2}+\frac{3}{2}+\ldots+\frac{n-1}{2}\right]\left(\frac{n}{2}\right) \text { terms } \\ &=\frac{1}{n} \cdot\left(\frac{n}{2}\right)^2 \quad\left[\because \text { sum of first } n \text { natural numbers }=n^2\right]\\ &=\frac{1}{n} \cdot \frac{n^2}{4}=\frac{n}{4} \\ \end{aligned}$$

Now, $\quad \sum x_i=1+2+3+4+\ldots+n=\frac{n(n+1)}{2}$

and $\quad \sum x_i^2=1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6}$

$$\begin{aligned} \therefore & =\sqrt{\frac{\sum x_i^2}{N}-\left(\frac{\sum x_i}{N}\right)^2} \\ & =\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^2(n+1)^2}{4 n^2}} \\ & =\sqrt{\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^2}{4}} \\ & =\sqrt{\frac{2\left(2 n^2+3 n+1\right)-3\left(n^2+2 n+1\right)}{12}} \\ & =\sqrt{\frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}} \\ & =\sqrt{\frac{n^2-1}{12}} \end{aligned}$$

Given,

$$\begin{aligned} & n_i=25, \bar{x}_i=18.2, \sigma_1=3.25 \\ & n_2=15, \sum_{i=1}^{15} x_i=279 \text { and } \sum_{i=1}^{15} x_i^2=5524 \end{aligned}$$

For first set, $$\sum x_i=25 \times 18.2=455$$

$\therefore \sigma_1^2=\frac{\Sigma x_i^2}{25}-(18.2)^2$

$$\begin{aligned} \Rightarrow \quad(3.25)^2 =\frac{\Sigma x_i^2}{25}-331.24 \\ \Rightarrow \quad 10.5625+331.24 =\frac{\Sigma x_i^2}{25} \\ \Rightarrow \quad \Sigma x_i^2 =25 \times(10.5625+331.24) \\ =25 \times 341.8025 \\ =8545.0625 \end{aligned}$$

For combined SD of the 40 observations $n=40$,

$$\begin{aligned} & \begin{aligned} \text { Now }\quad\Sigma x_i^2 & =5524+8545.0625=14069.0625 \\ \text { and } \quad\Sigma x_i & =455+279=734 \end{aligned} \\ & \qquad \begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{14069.0625}{40}-\left(\frac{734}{40}\right)^2} \\ & =\sqrt{351.726-(18.35)^2} \\ & =\sqrt{351.726-336.7225} \\ & =\sqrt{15.0035}=3.87 \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \text { Let } & x_i, i=1,2,3 \ldots, n_1 \text { and } y_j, j=1,2,3, \ldots, n_2 \\ \therefore & \bar{x}_1=\frac{1}{n_1} \sum_,\limits{i=1}^{n_1} x_i \text { and } \bar{x}_2=\frac{1}{n_2} \sum_\limits{i=1}^{n_2} y_j \\ \Rightarrow & \sigma_1^2=\frac{1}{n_1} \sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}_1\right)^2 \\ \text { and } & \sigma_2^2=\frac{1}{n_2} \sum_\limits{j=1}^n\left(y_j-\bar{x}_2\right)^2 \end{array}$$

Now, mean $\bar{x}$ of the given series is given by

$$\bar{x}=\frac{1}{n_1+n_2}\left[\sum_\limits{i=1}^{n_1} x_i+\sum_\limits{j=1}^{n_2} y_j\right]=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}$$

The variance $\sigma^2$ of the combined series is given by

$$\sigma^2=\frac{1}{n_1+n_2}\left[\sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}\right)^2+\sum_\limits{i=1}^{n_2}\left(y_j-\bar{x}\right)^2\right]$$

Now,

$$\begin{aligned} \sum_{i=1}^{n_1}\left(x_i-\bar{x}\right)^2 & =\sum_{i=1}^{n_1}\left(x_i-\bar{x}_j+\bar{x}_j-\bar{x}\right)^2 \\ & =\sum_{i=1}^{n_1}\left(x_i-\bar{x}_j\right)^2+n_1\left(\bar{x}_j-\bar{x}\right)^2+2\left(\bar{x}_j-\bar{x}\right) \sum_{i=1}^{n_1}\left(x_i-\bar{x}_j\right)^2 \end{aligned}$$

But $$\sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}_i\right)=0$$ [algebraic sum of the deviation of values of first series from their mean is zero]

Also, $$\sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}\right)^2=n_1 s_1^2+n_1\left(\bar{x}_1-\bar{x}\right)^2=n_1 s_1^2+n_1 d_1^2$$

Where, $$d_1=\left(\bar{x}_1-\bar{x}\right)$$

Similarly, $\quad \sum_{j=1}^{n_2}\left(y_j-\bar{x}\right)^2=\sum_{j=1}^{n_2}\left(y_j-\bar{x}_i+\bar{x}_i-\bar{x}\right)^2=n_2 s_2^2+n_2 d_2^2$

where, $d_2=\bar{x}_2-\bar{x}$

Combined SD, $\sigma=\sqrt{\frac{\left[n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)\right]}{n_1+n_2}}$

where, $d_1=\bar{x}_1-\bar{x}=\bar{x}_1-\left(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right)=\frac{n_2\left(\bar{x}_1-\bar{x}_2\right)}{n_1+n_2}$

and $d_2=\bar{x}_2-\bar{x}=\bar{x}_2-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}=\frac{n_1\left(\bar{x}_2-\bar{x}_1\right)}{n_1+n_2}$

$\therefore$ $\sigma^2=\frac{1}{n_1+n_2}\left[n_1 s_1^2+n_2 s_2^2+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}+\frac{n_2 n_1\left(\bar{x}_2-\bar{x}_1\right)^2}{\left(n_1+n_2\right)^2}\right]$

Also, $\sigma=\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}}$