Question

Find d for the arithmetic series with S17=-170 and a1=2

Answer 1

so, we know the sum of the first 17 terms is -170, thus S₁₇ = -170, and we also know the first term is 2, well


`\bf \textit{ sum of a finite arithmetic sequence}\\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ ----------\\ n=17\\ S_(17)=-170\\ a_1=2 \end{cases} \\\\\\ -170=\cfrac{17(2+a_(17))}{2}\implies \cfrac{-170}{17}=\cfrac{(2+a_(17))}{2} \\\\\\ -10=\cfrac{(2+a_(17))}{2}\implies -20=2+a_(17)\implies -22=a_(17)`

well, since the 17th term is that much, let's check what "d" is then anyway,


`\bf n^(th)\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ n=17\\ a_(17)=-22\\ a_1=2 \end{cases} \\\\\\ -22=2+(17-1)d\implies -22=2+16d\implies -24=16d \\\\\\ \cfrac{-24}{16}=d\implies -\cfrac{3}{2}=d`