A(n+1) = A(n)+4 <--- is a way to say, to get the next term, ADD 4 to the current one, whist A(1) = -2, is a way to say, the first term is -2.
[tex]\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}\\
----------\\
d=4\\
a_1=-2\end{cases}
\\\\\\
\begin{array}{llll}
term&value\\
------&------\\
2&a_2=-2+(2-1)4\\
3&a_3=-2+(3-1)4\\
4&a_4=-2+(4-1)4\\
5&a_5=-2+(5-1)4
\end{array}[/tex]
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A(n)= 1/4 * A(n) is just a way of saying, you get the next term by multiplying the current one by 1/4, and A(1) = 8, simply means the first term's value is 8.
[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=\frac{1}{4}\\
a_1=8
\end{cases}
\\\\\\
\begin{array}{llll}
term&value\\
------&------\\
2&a_2=8\left( \frac{1}{4} \right)^{2-1}\\\\
3&a_3=8\left( \frac{1}{4} \right)^{3-1}\\\\
4&a_4=8\left( \frac{1}{4} \right)^{4-1}\\\\
5&a_5=8\left( \frac{1}{4} \right)^{5-1}\\
\end{array}[/tex]
