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

1 answer

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:

\[ \boxed{-5} \]