Certainly! Let's look at the given geometric series to find the value of the first term [tex]\( a_1 \)[/tex]:
[tex]\[
\sum_{n=1}^{\infty} 12\left(-\frac{1}{9}\right)^{n-1}
\][/tex]
In a geometric series, the general form is:
[tex]\[
a + ar + ar^2 + ar^3 + \ldots
\][/tex]
Where [tex]\( a \)[/tex] is the first term ([tex]\( a_1 \)[/tex]) and [tex]\( r \)[/tex] is the common ratio.
Here, we're given a series in the form:
[tex]\[
12 \left(-\frac{1}{9}\right)^{n-1}
\][/tex]
To find the first term [tex]\( a_1 \)[/tex], we substitute [tex]\( n = 1 \)[/tex] into the expression. Doing so, we get:
[tex]\[
12 \left(-\frac{1}{9}\right)^{1-1}
\][/tex]
This simplifies to:
[tex]\[
12 \left(-\frac{1}{9}\right)^0
\][/tex]
Since any number raised to the power of 0 is 1, this further simplifies to:
[tex]\[
12 \cdot 1 = 12
\][/tex]
Therefore, the value of [tex]\( a_1 \)[/tex] is:
[tex]\[
\boxed{12}
\][/tex]
