To find the 10th term of a geometric sequence, we can use the general 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.

We are given:

• The 6th term, $a_6 = 30$,
• The common ratio, $r = 4$.

Step 1: Use the 6th term to find $a_1$

Substitute into the formula for the 6th term:

$a_6 = a_1 \cdot r^{6-1}$ $30 = a_1 \cdot 4^5$

Calculate $4^5$:

$4^5 = 1024$

Now, solve for $a_1$:

$30 = a_1 \cdot 1024$ $a_1 = \frac{30}{1024}$ $a_1 = \frac{15}{512}$

Step 2: Find the 10th term

Now that we have $a_1 = \frac{15}{512}$, we can find the 10th term using the formula:

$a_{10} = a_1 \cdot r^{10-1}$ $a_{10} = \frac{15}{512} \cdot 4^9$

First, calculate $4^9$:

$4^9 = 262144$

Now, substitute it into the formula:

$a_{10} = \frac{15}{512} \cdot 262144$

Multiply:

$a_{10} = \frac{15 \times 262144}{512} = \frac{3932160}{512} = 7680$

Final Answer:

The 10th term of the geometric sequence is 7680.

Would you like further details or explanations?

Related Questions:

1. What is the first term of the sequence?
2. How do you derive the formula for the $n$-th term in a geometric sequence?
3. What if the common ratio was negative, how would the 10th term change?
4. How would the solution differ if the 6th term was 60 instead of 30?
5. Can you find the sum of the first 10 terms of this sequence?

Tip:

Always calculate powers of the common ratio carefully, as errors in exponents can lead to incorrect results!