Alyssa eight numbers is formed by beginning with two given numbers each new number and the list is the product of the two previous numbers find the first number if the last three are shown 16, 64 and 1024

Let's denote the eight numbers as \( a_1, a_2, a_3, \ldots, a_8 \). The sequence is defined such that each new number is the product of the two previous numbers. The last three numbers provided are:

- \( a_6 = 16 \)
- \( a_7 = 64 \)
- \( a_8 = 1024 \)

We need to find the first number \( a_1 \).

Given the sequence pattern:

1. \( a_8 = a_7 \times a_6 \)
2. \( a_7 = a_6 \times a_5 \)
3. \( a_6 = a_5 \times a_4 \)

Starting from the given last three numbers:

1. \( a_8 = 1024 = a_7 \times a_6 = 64 \times 16 \)
2. \( a_7 = 64 = a_6 \times a_5 \)

From \( a_7 = 64 \) and \( a_6 = 16 \), we find \( a_5 \):

\[ a_5 = \frac{a_7}{a_6} = \frac{64}{16} = 4 \]

Next, using \( a_6 = 16 \) and \( a_5 = 4 \), we find \( a_4 \):

\[ a_4 = \frac{a_6}{a_5} = \frac{16}{4} = 4 \]

Continuing this process:

- \( a_5 = 4 \)
- \( a_4 = 4 \)
- \( a_3 = \frac{a_5}{a_4} = \frac{4}{4} = 1 \)
- \( a_2 = \frac{a_4}{a_3} = \frac{4}{1} = 4 \)
- \( a_1 = \frac{a_3}{a_2} = \frac{1}{4} = 0.25 \)

Thus, the first number \( a_1 \) is \( 0.25 \).

Would you like more details or have any questions?

Here are some questions you might be interested in:
1. How do we verify the calculations for each step?
2. Can you explain why each term is the product of the two previous terms?
3. What would happen if one of the starting numbers were zero?
4. How would the sequence change if we altered the formula?
5. Can you explain the significance of each number in the sequence?

**Tip:** Always verify your steps when solving problems to ensure accuracy.

Explore how to solve a sequence problem where each number is the product of the two previous numbers. Understand the sequence pattern and find the first number given the last three numbers in the sequence: 16, 64, and 1024. Suitable for students in Grades 9-12.

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