To solve for the seventeenth term of the geometric sequence, let's start by recalling the formula for the $n$th term of a geometric sequence:

$a_n = a_1 \cdot r^{(n-1)}$

where:

• $a_n$ is the $n$th term,
• $a_1$ is the first term,
• $r$ is the common ratio,
• $n$ is the term number.

Step 1: Identify the known values

Given:

• The first term, $a_1 = 1$,
• The tenth term, $a_{10} = 4$.

Step 2: Find the common ratio $r$

The formula for the tenth term is:

$a_{10} = a_1 \cdot r^{9}$

Substituting the known values:

$4 = 1 \cdot r^9$

$r^9 = 4$

Taking the ninth root of both sides to solve for $r$:

$r = \sqrt[9]{4}$

Step 3: Find the seventeenth term

Now, using the formula for the $n$th term, we want to find the seventeenth term, $a_{17}$:

$a_{17} = a_1 \cdot r^{16}$

Substituting the known values:

$a_{17} = 1 \cdot \left(\sqrt[9]{4}\right)^{16}$

Simplifying:

$a_{17} = \left(4^{\frac{16}{9}}\right)$

Final Answer:

The seventeenth term is:

$a_{17} = 4^{\frac{16}{9}} \approx 12.73$

This is the value of the seventeenth term.

Would you like more details or have any questions?

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Here are 5 related questions to explore further:

1. How do you find the sum of the first 17 terms of this geometric sequence?
2. What would be the common ratio if the first term was 2 instead of 1, keeping the tenth term as 4?
3. How does the geometric sequence change if the first term is negative?
4. What is the general formula for the sum of an infinite geometric series?
5. How would you solve for the 17th term if you were given a different geometric sequence?

Tip: Always remember that in geometric sequences, the common ratio $r$ determines how quickly the terms grow or shrink.