Statistics?

Let $S_{i}$ , be an element of the set $S$ . Because the elements of set $S$ form an arithmetic progression we have:

$S = \{A+d, A+2d, ... , A + (n-2)d, A + (n-1)d\}.$

We could set $A = 0$ because standard deviation is not affected by adding a constant to each of the elements of the set. To calculate we'll use the following formula:

${ \sigma }^{ 2 }=\left< x^{ 2 } \right> -{ \left< x \right> }^{ 2 }$

where,

$\left< x \right> =\frac { 1 }{ n } \sum { { x }_{ i } } =\frac { 1 }{ n } \sum _{ k=1 }^{ n }{ (k-1)d } \\ \qquad =\frac { 1 }{ n } (0+d+2d+...+(n-1)d)\\ \qquad =\frac { d }{ n } (1+2+3+...+(n-1));\quad 1+2+...+k=\frac { k(k+1) }{ 2 } \\ \qquad =\frac { d }{ n } \frac { (n-1)n }{ 2 } \\ \qquad =\frac { (n-1)d }{ 2 }$

and

$\left< { x }^{ 2 } \right> =\frac { 1 }{ n } \sum { { x }_{ i }^{ 2 } } =\frac { 1 }{ n } \sum _{ k=1 }^{ n }{ { (k-1) }^{ 2 }{ d }^{ 2 } } \\ \qquad =\frac { 1 }{ n } (0+{ d }^{ 2 }+4{ d }^{ 2 }+...+{ (n-1) }^{ 2 }{ d }^{ 2 })\\ \qquad =\frac { { d }^{ 2 } }{ n } (1+4+...+{ (n-1) }^{ 2 });\quad 1+{ 2 }^{ 2 }+{ 3 }^{ 2 }+...+{ k }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } \\ \qquad =\frac { { d }^{ 2 } }{ n } \frac { (n-1)(n)(2n-1) }{ 6 } \\ \qquad =\frac { { d }^{ 2 }(n-1)(2n-1) }{ 6 } \\ \qquad$ .

Plugging into the first equation:

${ \sigma }^{ 2 }=\frac { { d }^{ 2 }(n-1)(2n-1) }{ 6 } -\frac { { d }^{ 2 }{ (n-1) }^{ 2 } }{ 4 } \\ { \sigma }^{ 2 }=\frac { { d }^{ 2 }(n-1) }{ 12 } (2(2n-1)-3(n-1))\\ { \sigma }^{ 2 }=\frac { { d }^{ 2 }(n-1) }{ 12 } (4n-2-3n+3)\\ { \sigma }^{ 2 }=\frac { { d }^{ 2 }(n-1)(n+1) }{ 12 } \\ { \sigma }^{ 2 }=\frac { 3{ d }^{ 2 }({ n }^{ 2 }-1) }{ 36 } \\ \sigma =\frac { d }{ 6 } \sqrt { 3({ n }^{ 2 }-1) }$ .

That's it!