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Determine mean and standard deviation of first n terms of an A.P whose first term is 'a' and common difference d.

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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
$\qquad= \large\frac{d^2 (n^2-1)}{12}$
$\qquad= \sqrt { \large\frac{ d^2 (n^2-1)}{12}}$
$\qquad= d \sqrt {\large \frac{n^2-1}{12}}$
answered Jul 7, 2014 by meena.p

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