The arithmetic series of AP is $a_n= a+ (n-1)d$

$a_n =a+ (n-1) d$

Now, The arithematic series in two different ways .

Step 1:

$S_n =a_1 +(a_1+d) +(a_1+2d)+ .......(a_1+(n-2) d)+(a_1+(n-1) d)$----(1)

$S_n=(a_n-(n-1)d )+(a_n-(n-2)d)+.......(a_n-2d) +(a_n-d) +a_n$---(2)

Adding both sides of the equations

$2S_n=n (a_1+a_n)$

$S_n= \large\frac{n}{2} $$(a_1+a_n)$

$\therefore a_n= a_1 +(n-1)d$

$\therefore S_n= \large\frac{n}{2}$$ [a_1 +a_1 +(n-1)d]$

$\qquad=\large\frac{n}{2} $$ [ 2a_1+(n-1)d]$

Sum of the series= $\large\frac{n}{2}$$[2a_1+(n-1)d]$

Step 2:

$Mean =\large\frac{sum\;of\;the \;series}{n}$

$\qquad= \large\frac{\Large\frac{n}{2} [2a_1+(n-1) d]}{n}$

$\qquad= \large\frac{1}{2} $$[2a_1+(n-1)d]$

$\qquad= a_1+\large\frac{(n-1)}{2}$$d$

$S.D= \sqrt {variance}$

Variance = $\large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i }{n}\bigg)^2$

As $a_n = a_1+(n-1)d$

Where $i=0$ to $n$

$\sum x_i ^2= a_1^2+(a_1+d)^2+(a_1+2d)^2+.....(a_1+(n-1)d)$

$=> na_1^2+d^2(1+4+9+16+......(n-1)^2) + 2a_1d( 1+2+3+4+........(n-1))$

$=> \sum x_i ^2 =na_1^2+d^2 \bigg[ \large\frac {n(n-1)(2n-1)}{6} \bigg]+$$ 2a_1d \bigg[\large\frac{n(n-1)}{2} \bigg]$

$\large\frac{\sum x_i^2}{n}$$=a_1^2+\large\frac{d^2(n-1)(2n-1)}{6} $$+ a_1d (n-1)$

Variance = $\large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i }{n}\bigg)^2$

After solving the equation

$\qquad= \large\frac{d^2 (n^2-1)}{12}$

$\qquad= \sqrt { \large\frac{ d^2 (n^2-1)}{12}}$

$\qquad= d \sqrt {\large \frac{n^2-1}{12}}$