Harper Haynes
09/29/2023 · Elementary School
Qut Find the \( n^{\text {th }} \) and \( 10^{\text {th }} \) term of the geometric sequence \( 4,8,16,32, \ldots . \).
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Step-by-step Solution
To find the \( n^{\text{th}} \) term and the \( 10^{\text{th}} \) term of the geometric sequence \( 4, 8, 16, 32, \ldots \), we first identify the first term and the common ratio of the sequence.
1. **First term (\( a \))**: The first term \( a = 4 \).
2. **Common ratio (\( r \))**: The common ratio can be found by dividing the second term by the first term:
\[
r = \frac{8}{4} = 2.
\]
The formula for the \( n^{\text{th}} \) term of a geometric sequence is given by:
\[
a_n = a \cdot r^{n-1}.
\]
Now, substituting the values of \( a \) and \( r \):
\[
a_n = 4 \cdot 2^{n-1}.
\]
### Finding the \( n^{\text{th}} \) term:
The \( n^{\text{th}} \) term is:
\[
a_n = 4 \cdot 2^{n-1}.
\]
### Finding the \( 10^{\text{th}} \) term:
To find the \( 10^{\text{th}} \) term, we substitute \( n = 10 \) into the formula:
\[
a_{10} = 4 \cdot 2^{10-1} = 4 \cdot 2^9.
\]
Calculating \( 2^9 \):
\[
2^9 = 512.
\]
Now, substituting back:
\[
a_{10} = 4 \cdot 512 = 2048.
\]
### Summary:
- The \( n^{\text{th}} \) term is \( 4 \cdot 2^{n-1} \).
- The \( 10^{\text{th}} \) term is \( 2048 \).
Quick Answer
The \( n^{\text{th}} \) term is \( 4 \cdot 2^{n-1} \) and the \( 10^{\text{th}} \) term is \( 2048 \).
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