To find the number of terms in the given geometric series [tex]\(16 + 8 + 4 + \ldots + \frac{1}{16}\)[/tex], we need to determine the total count of terms from the first term to the last term.
Let's break this down step-by-step:
1. Identify the first term and common ratio:
- The first term of the geometric series is [tex]\(a = 16\)[/tex].
- Each term is half the previous term, so the common ratio [tex]\(r = \frac{1}{2}\)[/tex].
2. Determine the last term:
- The last term given in the series is [tex]\(\frac{1}{16}\)[/tex].
3. Formula for the [tex]\(n\)[/tex]-th term of a geometric series:
- The [tex]\(n\)[/tex]-th term [tex]\(T_n\)[/tex] of a geometric series is given by:
[tex]\[
T_n = a \cdot r^{(n-1)}
\][/tex]
- We know that [tex]\(T_n = \frac{1}{16}\)[/tex].
4. Set up the equation to find [tex]\(n\)[/tex]:
- Plug in the known values:
[tex]\[
\frac{1}{16} = 16 \cdot \left(\frac{1}{2}\right)^{(n-1)}
\][/tex]
5. Solve for [tex]\(n\)[/tex]:
- Simplify the equation:
[tex]\[
\frac{1}{16} = 16 \cdot \left(\frac{1}{2}\right)^{(n-1)}
\][/tex]
- Divide both sides by 16:
[tex]\[
\frac{1}{16} \div 16 = \left(\frac{1}{2}\right)^{(n-1)}
\][/tex]
[tex]\[
\frac{1}{256} = \left(\frac{1}{2}\right)^{(n-1)}
\][/tex]
- Recognize that [tex]\(\frac{1}{256} = \left(\frac{1}{2}\right)^8\)[/tex]:
[tex]\[
\left(\frac{1}{2}\right)^8 = \left(\frac{1}{2}\right)^{(n-1)}
\][/tex]
- Since the bases are the same, the exponents must be equal:
[tex]\[
n - 1 = 8
\][/tex]
- Solve for [tex]\(n\)[/tex]:
[tex]\[
n = 8 + 1 = 9
\][/tex]
Therefore, the number of terms in the geometric series is [tex]\(\boxed{9}\)[/tex].
