Question

Find the 96th term of the arithmetic sequence 1,−12,−25

Answer 1

`1~~,~~\stackrel{1-13}{-12}~~,~~\stackrel{-12-13}{-25}~~,~~...\hspace{5em}\stackrel{\textit{common difference}}{-13} \\\\[-0.35em] ~\dotfill\\\\ 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}\\[-0.5em] \hrulefill\\ a_1=1\\ d=-13\\ n=96 \end{cases} \\\\\\ a_(96)=1+(96-1)(-13)\implies a_(96)=1+(-1235)\implies \boxed{a_(96)=-1234}`