5 + 9 + 13 + ... + tn = 945

We need the sum formula:

Sn = [2t1 + (n-1)·d]

To find d, the common difference, we subtract the first term,
t1 = 5 from the second term t2 = 9, and
get 9-5 = 4, and as a check we also subtract the second term,
t2 = 9 from the third term t3 = 12, and
get 13-9 = 4, so the common difference d is 4.  So we substitute
d = 4, Sn = 945, t1 = 5 and solve for n:

Sn = [2t1 + (n-1)·d]
945 = [2·5 + (n-1)·4]

Multiply both sides by 2 to clear the fraction on the right:

1890 = n[10 + (n-1)·4]

1890 = n[10 + 4(n-1)]

1890 = n[10 + 4n - 4]

1890 = n[4n + 6]

1890 = 4n² + 6n

Divide through by 2

945 = 2n² + 3n

Get 0 on the left

0 = 2n² + 3n - 945

The right side factors as

0 = (n - 21)(2n + 45)

Use the zero-factor property:

    n - 21 = 0;   2n + 45 = 0
         n = 21;       2n = -45
                        n = 

Ignore the negative answer.

It has 21 terms.

5 + 9 + 13 + 17 + 21 + 25 + 29 + 33 + 37 + 41 + 45 + 49 + 53 + 57 + 61 + 65 + 69 + 73 + 77 + 81 + 85

Edwin