Find the nth term of this quadratic sequence:
4, 7, 12, 19, 28

n² + 3

Step-by-step explanation:

It's a quadratic sequence, so it follows the form:

y = ax² + bx + c

We're given five points that satisfy the equation.  (1, 4), (2, 7), (3, 12), (4, 19), and (5, 28).  Picking any three points, we can form a system of equations.

If we pick (1, 4), (2, 7), and (4, 19):

4 = a(1)² + b(1) + c

7 = a(2)² + b(2) + c

19 = a(4)² + b(4) + c

4 = a + b + c

7 = 4a + 2b + c

19 = 16a + 4b + c

Through substitution, elimination, or trial and error, we can find a = 1, b = 0, and c = 3.

y = x² + 3

So the nth term of the sequence is n² + 3.

a(n) = a(n-1) + (2n - 1)

Step-by-step explanation:

Start by analyzing the pattern:

7 is 3 more than 4;

12 is 5 more than 7;

19 is 7 more than 12, and so on.  

Each step is an odd number and is 2 greater than the previous step.

a(2) = 7 = 4 + step = 4 + 3 = 7

a(3) = 12 = 7 + step = 7 + 5 = 12

a(4) =                      12 + 7  = 19

a(5) =                       19 + 9 = 28

and so on.  

Looking at a(2), we see that the step is 2+1, or 3;

Looking at a(3), we see that the step is 2(3) - 1, or 5;

Looking at a(4), we see that the step is 2(4) - 1, or 7; and so on.

Looking at a(n), we see that the step is 2n - 1.

Thus, a(n) = a(n-1) + (2n - 1)

7

step-by-step explanation:

i think the answer would be c