Question

What the recursive formula for the sequence

Answer 1

`\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} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})`

so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

`\bf \begin{cases} a_1=2\\ a_n=a_(n-1)-5 \end{cases}`

just a quick note on notation:

`\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_(n-1)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_(5-1)\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_(12)}\qquad \quad \stackrel{\textit{term before it}}{a_(12-1)\implies a_(11)}`