Answer please! Find the answer to both questions

APtₙ = (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 + 1k² + 3k + 3 - (k + 2) =  k² + 3k + 3 - k - 2 = k² + 2k + 12k² + 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

a₁= -k² - k + 1

Common difference is

d = k² + 2k + 1

nth term is:

tₙ = a₁ + (n-1)dtₙ = -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

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