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**Hint:**An arithmetic series is the sum of a sequence $\left\{a_{k}\right\}, k=1,2, \ldots,$ in which each term is computed from the previous one by adding (or subtracting) a constant d. Therefore, for $k>1$, $a_{k}=a_{k-1}+d=a_{k-2}+2 d=\ldots=a_{1}+d(k-1)$

The sum of the sequence of the first n terms is then given by:

${{S}_{n}}\text{ }\equiv \sum\limits_{k=1}^{n}{{{a}_{k}}}=\sum\limits_{k=1}^{n}{\left[ {{a}_{1}}+(k-1)d \right]}$

$\mathrm{Sn}=\mathrm{n}(\mathrm{a} 1+\mathrm{an}) 2$

$=n{{a}_{1}}+d\sum\limits_{k=1}^{n-1}{k}$

**Complete step-by-step answer:**An arithmetic sequence is a sequence where the difference d between successive terms is constant. An arithmetic series is the sum of the terms of an arithmetic sequence. The nth partial sum of an arithmetic sequence can be calculated using the first and last terms as follows:

$\mathrm{Sn}=\mathrm{n/2}(\mathrm{a_1} +\mathrm{a_n}) $

An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference. A

geometric sequence is a sequence with the ratio between two consecutive terms

constant.

Given series is $5+11+17+\ldots+95$.

Thus, $a=5, d=11-5=6, l=95$

$\mathrm{n}=\dfrac{\mathrm{l}-\mathrm{a}}{\mathrm{d}}+1$

$=\dfrac{95-5}{6}+1$

$=\dfrac{90}{6}+1=15+1=16$

Therefore, $\mathrm{S}_{\mathrm{n}}=\dfrac{\mathrm{n}}{2}(\mathrm{a}+\mathrm{l})$

$=\dfrac{16}{2}(5+95)$

$=8(100)$

$=800$

**Note:**Using the sum identity

$\sum_{k=1}^{n} k=\dfrac{1}{2} n(n+1)$ then gives $S_{n}=n a_{1}+\dfrac{1}{2} d n(n-1)=\dfrac{1}{2} n\left[2 a_{1}+d(n-1)\right]$

Note, however, that $a_{1}+a_{n}=a_{1}+\left[a_{1}+d(n-1)\right]=2 a_{1}+d(n-1)$

So $S_{n}=\dfrac{1}{2} n\left(a_{1}+a_{n}\right)$ or n times the arithmetic mean of the first and last terms.

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