Question

I'm in 11th grade algebra, problem below, thank you!

Answer 1

$Answer: 2+(-1)+(-4)+...+(-238)=-9558.\\\sum\limits_{k=1}^{89}(5k+1)=20114.$

Step-by-step explanation:

$2+(-1)+(-4)+...+(-238)=\\a_1=2\ \ \ \ \ a_n=-238.\\d=(-1)-2\\d=-3.\\a_n=a_1+(n-1)*d\\-238=2+(n-1)*(-3)\\-238=2-3n+3\\-238=5-3n\\-243=-3n\\-3*81=-3*n\\Divide\ both \ sides\ of\ the\ equation\ by\ -3:\\81=n.$

                   $\diplaystyle\\\\\boxed {S=\frac{a_1+a_n}{2}*n }$

 $\displaystyle\\S=\frac{2+(-238)}{2} *81\\S=\frac{-236}{2}*81\\ S=-118*81\\S=-9558.$

$\displaystyle\\\sum\limits_{k=1}^{89}(5k+1)=\\a_1=5*1+1\\a_1=5+1\\a_1=6\\a_2=5*2+1\\a_2=10+1\\a_2=11.\\d=a_2-a_1\\d=11-6\\d=5.$

                   $\displaystyle\\\boxed {S=\frac{2a_1+(n-1)*d}{2}*n }$  

$\displaystyle\\S=\frac{2*6+(89-1)*5}{2}*89\\\\ S=\frac{12+88*5}{2} *89\\\\S=\frac{12+440}{2}*89\\\\ S=\frac{452}{2} *89\\\\S=226*89\\\\S=20114.$