Answer:
- AP
- tₙ = (n-2)k² + (2n - 3)k + n
Step-by-step explanation:
Given sequence:
- -k² - k + 1, k + 2, k² + 3k + 3, 2k² + 5k + 4, ...
We can see apart from the second term all terms are 2nd degree trinomial so it is not a GP.
Let's find the difference: to determine if it is a AP:
- k + 2 - (-k² - k + 1) = k + 2 + k² + k -1 = k² + 2k + 1
- k² + 3k + 3 - (k + 2) = k² + 3k + 3 - k - 2 = k² + 2k + 1
- 2k² + 5k + 4 - (k² + 3k + 3) = 2k² + 5k + 4 - k² - 3k - 3 = k² + 2k + 1
As we see the difference is common and it confirms the sequence is AP.
First term is
Common difference is
nth term is:
- tₙ = a₁ + (n-1)d
- tₙ = -k² - k + 1 + (n-1)(k² + 2k + 1) =
- -k² - k + 1 + (n-1)k² + (n-1)*2k + n-1 =
- (n-2)k² + (2n - 3)k + n
