Find the 69th term in the sequence–10, –5, 2, 11, 22, 35, . . . . .

This sequence appears to be arithmetic, meaning that the difference between consecutive terms is constant. Let's find the common difference:

-5 - (-10) = 5 2 - (-5) = 7 11 - 2 = 9 22 - 11 = 11 35 - 22 = 13

The common difference is not constant, so this is not an arithmetic sequence. However, the differences between the terms are increasing by 2 each time. This suggests that the sequence is quadratic, meaning it can be represented by a second-degree polynomial.

Let's denote the first term of the sequence as a, the second term as b, the third term as c, and so on. We can then write the following equations based on the differences between the terms:

b - a = 5 c - b = 7 d - c = 9 e - d = 11 f - e = 13

We can see that the difference between consecutive terms is increasing by 2 each time. This suggests that the second difference is constant, which is a characteristic of quadratic sequences.

The nth term of a quadratic sequence can be represented as an^2 + bn + c. To find the coefficients a, b, and c, we can set up a system of equations using the first three terms of the sequence:

a + b + c = -10 4a + 2b + c = -5 9a + 3b + c = 2

Solving this system of equations, we find that a = 2, b = -7, and c = -5. Therefore, the nth term of the sequence is 2n^2 - 7n - 5.

To find the 69th term, we substitute n = 69 into this equation:

2(69)^2 - 7(69) - 5 = 9383.