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We know that \(a_6 = 2a_5 - a_4\text{.}\) So to find \(a_6\) we need to find \(a_5\) and \(a_4\text{.}\) Well
\begin{equation*}
a_5 = 2a_4 - a_3 \qquad \text{and} \qquad a_4 = 2a_3 - a_2\text{,}
\end{equation*}
so if we can only find \(a_3\) and \(a_2\) we would be set. Of course
\begin{equation*}
a_3 = 2a_2 - a_1 \qquad \text{and} \qquad a_2 = 2a_1 - a_0\text{,}
\end{equation*}
so we only need to find \(a_1\) and \(a_0\text{.}\) But we are given these. Thus
\begin{align*}
a_0 \amp = 3\\
a_1 \amp = 4\\
a_2 \amp = 2\cdot 4 - 3 = 5\\
a_3 \amp = 2\cdot 5 - 4 = 6\\
a_4 \amp = 2\cdot 6 - 5 = 7\\
a_5 \amp = 2\cdot 7 - 6 = 8\\
a_6 \amp = 2\cdot 8 - 7 = 9\text{.}
\end{align*}
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