Answer :
To determine the value of [tex]\( a_1 \)[/tex] for the given geometric series
[tex]\[ \sum_{n=1}^{\infty} 12\left(-\frac{1}{9}\right)^{n-1}, \][/tex]
let's break down the series into its components.
A geometric series is of the form:
[tex]\[ \sum_{n=1}^{\infty} a_1 \cdot r^{n-1}, \][/tex]
where:
- [tex]\( a_1 \)[/tex] is the first term,
- [tex]\( r \)[/tex] is the common ratio.
In the given series:
[tex]\[ \sum_{n=1}^{\infty} 12\left(-\frac{1}{9}\right)^{n-1}, \][/tex]
we can identify the first term [tex]\( a_1 \)[/tex] and the common ratio [tex]\( r \)[/tex].
Here, the first term [tex]\( a_1 \)[/tex] is the coefficient outside the exponentiated term, which is 12.
So,
[tex]\[ a_1 = 12. \][/tex]
Thus, the value of [tex]\( a_1 \)[/tex] is:
12
[tex]\[ \sum_{n=1}^{\infty} 12\left(-\frac{1}{9}\right)^{n-1}, \][/tex]
let's break down the series into its components.
A geometric series is of the form:
[tex]\[ \sum_{n=1}^{\infty} a_1 \cdot r^{n-1}, \][/tex]
where:
- [tex]\( a_1 \)[/tex] is the first term,
- [tex]\( r \)[/tex] is the common ratio.
In the given series:
[tex]\[ \sum_{n=1}^{\infty} 12\left(-\frac{1}{9}\right)^{n-1}, \][/tex]
we can identify the first term [tex]\( a_1 \)[/tex] and the common ratio [tex]\( r \)[/tex].
Here, the first term [tex]\( a_1 \)[/tex] is the coefficient outside the exponentiated term, which is 12.
So,
[tex]\[ a_1 = 12. \][/tex]
Thus, the value of [tex]\( a_1 \)[/tex] is:
12
