Solution:
Given that the first 
Given AP in tabular form is shown below,
![\[\begin{tabular}{|l|l|l|} \hline$x_{i}$ & $d_{i}=x_{i}-a$ & $d_{i}^{2}$ \\ \hline$a$ & 0 & 0 \\ \hline$a+d$ & $d$ & $d^{2}$ \\ \hline$a+2 d$ & $2 d$ & $4 d^{2}$ \\ \hline$a+3 d$ & $3 d$ & $9 d^{2}$ \\ \hline$-$ & $-$ & $-$ \\ \hline$a+(n-1) d$ & $(n-1) d$ & $(n-1)^{2} d^{2}$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-583e5e52b26d32788d27cb3bd749cd72_l3.png "Rendered by QuickLaTeX.com")
Assumed 
Given that the AP have
![\sum \mathrm{x}_{\mathrm{i}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ccacd7fdca7cae8d8d155df113e2e034_l3.png "Rendered by QuickLaTeX.com")
Calculate the actual mean,
Substitute the corresponding values,
![\bar{x}=\frac{\frac{n}{2}[2 a+(n-1) d]}{n}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c2603cb30b63379a0a0aa207a9987b16_l3.png "Rendered by QuickLaTeX.com")
Above eq. can be written as
![\begin{aligned} &\bar{x}=\frac{[2 a+(n-1) d]}{2} \\ &\bar{x}=a+\frac{[(n-1) d]}{2} \\ &\bar{x}=a+\frac{(n-1)}{2} d \end{aligned}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31e945e1790366c7153507d8a0d547b0_l3.png "Rendered by QuickLaTeX.com")
The other two columns sums, i.e. 
![\sum d_{i}=\sum\left(x_{i}-a\right)=d[1+2+3+\cdots+(n-1)]=d\left(\frac{n(n-1)}{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ea82e5d921f1061345ff60f95b4fa6db_l3.png "Rendered by QuickLaTeX.com")
![\sum \mathrm{d}_{\mathrm{i}}^{2}=\sum\left(\mathrm{x}_{\mathrm{i}}-\mathrm{a}\right)^{2}=\mathrm{d}^{2}\left[1^{2}+2^{2}+3^{2}+\cdots+(\mathrm{n}-1)^{2}\right]=\mathrm{d}^{2}\left(\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9583ea4fc094a02d038ed30dd198c4d9_l3.png "Rendered by QuickLaTeX.com")
The standard deviation is given by,
Substitute the corresponding values,
Cancel the like terms,
Take out the common terms
Taking the LCM,
As a result the mean and standard deviation of the given AP is
