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

Answer

N/a

Explanation

`because Mean=(Sigmax_(i))/n=1/n[n/2(2a+(n-1))d]
`=a+((n-1))/(2)d`
`therefore Sigma(x_(i)-a)=d[1+2+3+_...+(n-1)d]`
=d ((n-1))/2`
and `Sigma(x_(i)-a)^(2)=d^(2)[1^(2)+2^(2)+3^(2)+...+(n-1)^(2)]`
`=(d^(2)(n-1)n(2n-1))/6`
`=dsqrt(((n-1)(2n-1))/6-((n-1)^(2))/(4))`
`dsqrt(((n-1))/(2)((2n-1)/(3)-(n-1)/2))`
`=dsqrt(((n-1))/2[(4n-2-3n+3)/(6)]`
`dsqrt(((n-1)(n+1))/12)=dsqrt(((n^(2)-1))/(12)`