Question

Question #e4948

Answer 1

`970`

Explanation:

For the standard `color(blue)"arithmetic sequence"`

`a,a+d,a+2d,a+3d,......,a+(n-1)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= ....=a_n-a_(n-1)`

`color(blue)"The sum to n terms" = color(red)(bar(ul(|color(white)(2/2)color(black)(S_n=n/2[2a+(n-1)d])color(white)(2/2)|)))`

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

Here `a=1, d=6-1=11-6=5" and " n=20`

`rArrS_20=20/2[(2xx1)+(19xx5)]`

`=10(2+95)=970`

Answer 2

`970`

Explanation:

We have: `1+6+11+...+96`

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

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

`=> T_(n) = T_(1) + (n - 1) d`

`=> 96 = 1 + (n - 1) (6 - 1)`

`=> 95 = 5 (n - 1)`

`=> n - 1 = 19`

`therefore n = 20`

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

`=> S_(20) = (20) / (2) (1 + 96)`

`=> S_(20) = 10 cdot 97`

`therefore S_(20) = 970`