Evaluate arithmetic series:-
Step-by-step answer, please!

Let's see
[tex]\\ \rm\Rrightarrow {\displaystyle{\sum^{275}_{k=1}}}(-5k+12)[/tex]
[tex]\\ \rm\Rrightarrow (-5(1)+12)+(-5(2)+12)\dots (-5(275)+12)[/tex]
[tex]\\ \rm\Rrightarrow 7+5+3+2+1+\dots -1363[/tex]
So
Sum:-
[tex]\\ \rm\Rrightarrow S_n=\dfrac{n}{2}[a+l][/tex]
[tex]\\ \rm\Rrightarrow \dfrac{275}{2}(7-1363)[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{275}{2}(-1356)[/tex]
[tex]\\ \rm\Rrightarrow 275(-678)[/tex]
[tex]\\ \rm\Rrightarrow -186450[/tex]
Answer:
-186,450
Step-by-step explanation:
Sum of arithmetic series formula
[tex]S_n=\dfrac{n}{2}[2a+(n-1)d][/tex]
where:
[tex]\displaystyle \sum\limits_{k=1}^{275} (-5k+12)[/tex]
To find the first term, substitute [tex]k = 1[/tex] into [tex](-5k+12)[/tex]
[tex]a_1=-5(1)+12=7[/tex]
To find the common difference, find [tex]a_2[/tex] then subtract [tex]a_1[/tex] from [tex]a_2[/tex]:
[tex]a_2=-5(2)+12=2[/tex]
[tex]\begin{aligned}d & =a_2-a_1\\ & =2-7\\ & =-5\end{aligned}[/tex]
Given:
[tex]\begin{aligned}S_{275} & = \dfrac{275}{2}[2(7)+(275-1)(-5)]\\& = \dfrac{275}{2}[14-1370]\\& = \dfrac{275}{2}[-1356]\\& = -186450\end{aligned}[/tex]