Find the nth term of 11, 20, 35, 56, 83, . . .

nth number is 83 hope you understand my answer thanks

3n^2 + 8

Step-by-step explanation:

an^2 + bn + c

It is a quadratic sequence

11, 20, 35, 56, 83

 +8 +15 +21 +27

    +6   +6   +6

a + b + c

4a + 2b + c

9a + 3b + c

4a + 2b + c - a + b + c = 20 - 11

4a + 2b + c - a + b + c = 9

3a + b = 9

9a + 3b + c - 4a + 2b + c = 35 - 20

5a + b = 15

-2a = -6

a = -6/-2

a = 3

Substitute to find the values of b & c

For b

5(3) + b = 15

15 + b = 15

b = 15 - 15

b = 0

For c

9(3) + 3(0) + c = 35

27 + c = 35

c = 35 - 27

c = 8

3n^2 + 0n + 8

3n^2 + 8

its 4 because 2 times 2 is 4, if you double the number two you get four

what's the question?

step-by-step explanation:

there is no question.