Answer
$a_n=152-14n$
Work Step by Step
Let $a_n$ be our arithmetic sequence. We are given:
$$\begin{align*}
a_4&=96\\
d&=-14.
\end{align*}$$
We determine the first term $a_1$:
$$\begin{align*}
a_n&=a_1+(n-1)d\\
a_4&=a_1+(4-1)d\\
96&=a_1+3(-14)\\
96+42&=a_1\\
a_1&=138.
\end{align*}$$
We write the rule for the general term $a_n$:
$$\begin{align*}
a_n&=a_1+(n-1)d\\
a_n&=138+(n-1)(-14)=138-14n+14=152-14n.
\end{align*}$$
We got:
$$a_n=152-14n.\tag1$$
We calculate the first six terms substituting $n=2,3,5,6$ in Eq. $(1)$ and $a_1=138$, $a_4=96$:
$$\begin{align*}
a_1&=138\\
a_2&=152-14(2)=124\\
a_3&=152-14(3)=110\\
a_4&=96\\
a_5&=152-14(5)=82\\
a_6&=152-14(6)=68.
\end{align*}$$
Graph the first six terms:
