Question

Use the recursively defined geometric sequence a1 = ratio. (1 point)
10
3
6, an = 4an-1 and find the common
2
3
5
6
4

Answer 1

We are given a geometric sequence defined recursively where the first term \( a_1 = 10 \) and the recursive relation \( a_n = 4a_{n-1} \).

To find the next few terms of this sequence:

1. **Calculate \( a_2 \)**:
\[
a_2 = 4a_1 = 4 \times 10 = 40
\]

2. **Calculate \( a_3 \)**:
\[
a_3 = 4a_2 = 4 \times 40 = 160
\]

3. **Calculate \( a_4 \)**:
\[
a_4 = 4a_3 = 4 \times 160 = 640
\]

4. **Calculate \( a_5 \)**:
\[
a_5 = 4a_4 = 4 \times 640 = 2560
\]

So the first five terms of the sequence are:
- \( a_1 = 10 \)
- \( a_2 = 40 \)
- \( a_3 = 160 \)
- \( a_4 = 640 \)
- \( a_5 = 2560 \)

The common ratio can be calculated as:
\[
\text{Common Ratio} = \frac{a_n}{a_{n-1}} = 4
\]

This matches the recursive rule \( a_n = 4a_{n-1} \).

If there are any additional questions or details you'd like to explore, please let me know!