Question

Whats the common difference of a^1 = 4, a^n = a^n-1 - 5

Answer 1

It appears that you're describing a sequence defined recursively. The formula you provided states that the first term \( a^1 = 4 \), and for \( n \geq 2 \), each term \( a^n \) is defined as follows:

\[
a^n = a^{n-1} - 5
\]

To find the common difference, let's calculate the first few terms of the sequence:

1. **First term**: \( a^1 = 4 \)
2. **Second term**:
\[
a^2 = a^{1} - 5 = 4 - 5 = -1
\]
3. **Third term**:
\[
a^3 = a^{2} - 5 = -1 - 5 = -6
\]
4. **Fourth term**:
\[
a^4 = a^{3} - 5 = -6 - 5 = -11
\]

Now we have the first four terms of the sequence: \( 4, -1, -6, -11 \).

To find the common difference, we subtract consecutive terms:

- From \( a^1 \) to \( a^2 \): \( -1 - 4 = -5 \)
- From \( a^2 \) to \( a^3 \): \( -6 - (-1) = -5 \)
- From \( a^3 \) to \( a^4 \): \( -11 - (-6) = -5 \)

The difference between each consecutive term is the same, which is \( -5 \).

Therefore, the common difference of the sequence is:

\[
\boxed{-5}
\]