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# What is the 32nd term of the arithmetic sequence where a_{1} = -33 and a_{9} = -121?

**Solution:**

The nth term of an arithmetic sequence whose first term is a_{1} and common difference is d is given by:

a_{1} + (n - 1) d

It is given that

a1 = -33 and a_{9} = -121

We know that

a_{9} = a_{1} + (9 - 1) d

-33 + 8d = - 121

8d = -121 + 33

8d = -88

d = -88/8

d = -11

Now we have to find the 32nd term

a_{32} = a1 + (32 - 1) d

Substituting the values

a_{32} = -33 + (31) (-11)

a_{32} = -33 - 341

So we get

a_{32} = -374

Therefore, the 32nd term is -374.

## What is the 32nd term of the arithmetic sequence where a_{1} = -33 and a_{9} = -121?

**Summary:**

The 32nd term of the arithmetic sequence where a_{1} = -33 and a_{9} = -121 is -374.

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