What is a35 for the arithmetic sequence presented in the table below? n 5 10 an 32 7 hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference. a.) a35 = −98 b.) a35 = −118 c.) a35 = −112 d.) a35 = −102

choice B) a35 = -118

Explanation:

When n = 5, an = 32 as shown in the first column of the table. This means the fifth term is 32. Plug in those values to get

an = a1+d(n-1)

32 = a1+d(5-1)

32 = a1+4d

Solve for a1 by subtracting 4d from both sides

a1 = 32-4d

We'll plug this in later

Turn to the second column of the table. We have n = 10 and an = 7. Plug those values into the formula

an = a1+d(n-1)

7 = a1 + d(10-1)

7 = a1+9d

Now substitute in the equation in which we solved for a1

7 = a1+9d

7 = 32-4d+9d ... replace a1 with 32-4d

7 = 32+5d

5d = 7-32

5d = -25

d = -25/5

d = -5

This tells us that we subtract 5 from each term to get the next term.

Use this d value to find a1

a1 = 32-4d

a1 = 32-4*(-5)

a1 = 32+20

a1 = 52

The first term is 52

The nth term formula is therefore

an = 52 + (-5)(n-1)

which simplifies to

an = -5n + 57

To check this result, plug in n = 5 to find that a5 = 32. Similarly, you'll find that a10 = 7 after plugging in n = 10. I'll let you do these checks.

Replace n with 35 to find the 35th term

an = -5n + 57

a35 = -5(35) + 57

a35 = -175 + 57

a35 = -118

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.