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.`

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=(a_1+a_n)/(2)*n }`

`\displaystyle\\S=(2+(-238))/(2) *81\\S=(-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=(2a_1+(n-1)*d)/(2)*n }`

`\displaystyle\\S=(2*6+(89-1)*5)/(2)*89\\\\ S=(12+88*5)/(2) *89\\\\S=(12+440)/(2)*89\\\\ S=(452)/(2) *89\\\\S=226*89\\\\S=20114.`