# Question #e4948

Dec 18, 2016

$970$

#### Explanation:

For the standard $\textcolor{b l u e}{\text{arithmetic sequence}}$

$a , a + d , a + 2 d , a + 3 d , \ldots \ldots , a + \left(n - 1\right) d$

where a$= {a}_{1}$ is the first term, d the common difference and n the number of terms.

and $d = {a}_{2} - {a}_{1} = {a}_{3} - {a}_{2} = \ldots . = {a}_{n} - {a}_{n - 1}$

$\textcolor{b l u e}{\text{The sum to n terms}} = \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

A series is the sum of the terms in the sequence.

Here $a = 1 , d = 6 - 1 = 11 - 6 = 5 \text{ and } n = 20$

$\Rightarrow {S}_{20} = \frac{20}{2} \left[\left(2 \times 1\right) + \left(19 \times 5\right)\right]$

$= 10 \left(2 + 95\right) = 970$

Dec 18, 2016

$970$

#### Explanation:

We have: $1 + 6 + 11 + \ldots + 96$

This is an arithmetic sequence with a common difference of $7$.

First, let's determine the number of terms in the sequence:

$\implies {T}_{n} = {T}_{1} + \left(n - 1\right) d$

$\implies 96 = 1 + \left(n - 1\right) \left(6 - 1\right)$

$\implies 95 = 5 \left(n - 1\right)$

$\implies n - 1 = 19$

$\therefore n = 20$

Then, let's evaluate the sum of the $20$ terms of this arithmetic sequence:

$\implies {S}_{20} = \frac{20}{2} \left(1 + 96\right)$

$\implies {S}_{20} = 10 \cdot 97$

$\therefore {S}_{20} = 970$