Question

Find a1, for the arithmetic series with S14 = - 420 and d = -6.

Answer 1

$\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} n = 14\\ d= -6 \end{cases} \\\\\\ a_{14}=a_1+(14-1)(-6)\implies a_{14}=a_1+(13)(-6)\implies a_{14}=a_1-78 \\\\[-0.35em] ~\dotfill$

$\bf \textit{sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ \cline{1-1} n= 14\\ S_{14}=-420\\ a_{14}=a_1-78 \end{cases}\implies S_{14}=\cfrac{n(a_1+a_{14})}{2} \\\\\\ -420=\cfrac{14[a_1+(a_1-78)]}{2}\implies -420=7(2a_1-78)\implies \cfrac{-420}{7}=2a_1-78 \\\\\\ -60=2a_1-78\implies 18=2a_1\implies \cfrac{18}{2}=a_1\implies 9=a_1$