Question

Find n for the sequence for which a1 = 30, d = -4 and sn= -210.

Answer 1

I must say the reply above is superb, lemme post here, hopefully not sounding too redundant.

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

now let's plug that for aₙ in the summation formula

$\bf \textit{ sum of a finite arithmetic sequence}\\\\S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\----------\\a_1=30\\a_n=34-4n\\S_n=-210\end{cases}\\\\\\-210=\cfrac{n(30+34-4n)}{2}\implies -420=n(64-4n)\\\\\\-420=64n-4n^2\implies \stackrel{\textit{dividing by 4}}{-105=16n-n^2}\\\\\\n^2-16n-105=0\implies (n-21)(n+5)=0\implies n=\begin{cases}\boxed{21}\\-5\end{cases}$

Answer 2

$n\in\mathbb{N^+}$

The formula of the sum of an arithmetic sequence:

$S_n=\dfrac{2a_1+(n-1)d}{2}\cdot n$

We have:

$a_1=30;\ d=-4;\ S_n=-210$

substitute:

$\dfrac{2\cdot30+(n-1)(-4)}{2}\cdot n=-210\ \ \ |\cdot2\\\\(60-4n+4)n=-420\\\\(64-4n)n=-420\\\\64n-4n^2=-420\ \ \ |:(-4)\\\\n^2-16n=105\ \ \ |-105\\\\n^2-16n-105=0\\\\n^2+5n-21n-105=0\\\\n(n+5)-21(n+5)=0\\\\(n+5)(n-21)+0\iff n+5=0\ \vee\ n-21=0\\\\n=-5\notin\mathbb{N^+}\ \vee\ n=21\in\mathbb{N^+}$

Answer: b. 21