Question

HELP PLEASE

Find n for a series for which a1 = 30, d = -4, and Sn = -210.
a.
-21
c.
-10
b.
21
d.
10

Answer 1

$\bf \qquad \qquad \textit{sum of a finite arithmetic sequence}\\\\ S_n=\cfrac{n}{2}(a_1+a_n)~ \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ a_n=n^{th}~value\\ ----------\\ a_1=30\\ S_n=-210 \end{cases} \implies -210=\cfrac{n}{2}(30+a_n)\\\\\\ -420=n(30+a_n)\implies -420=30n+na_n \\\\\\ -420-30n=na_n\implies \boxed{\cfrac{-420-30n}{n}=a_n}\\\\ -------------------------------$

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=30\\ d=-4 \end{cases} \\\\\\ a_n=30+(n-1)(-4)\implies a_n=30-4n+4 \\\\\\ \boxed{a_n=34-4n}\\\\ -------------------------------$

$\bf \stackrel{a_n}{34-4n}~~=~~\stackrel{a_n}{\cfrac{-420-30n}{n}}\implies 34n-4n^2=-420-30n \\\\\\ 0=4n^2-64n-420\implies 0=4(n^2-16n-105) \\\\\\ 0=n^2-16n-105\implies 0=(n-21)(n+5)\implies n= \begin{cases} \boxed{21}\\ -5 \end{cases}$

since the term is just a term unit in the sequence, it cannot be a negative value, thus is not -5.