# What is the 32nd term of the arithmetic sequence where a_{1} = -31 and a_{9} = -119?

**Solution:**

The n^{th} term of an arithmetic sequence is given by a_{n} = a + (n - 1)d.

Given the first term a = a_{1} = -31

common difference = d = unknown

To find the common difference substitute in the formula, we get

a_{9} = -119

⇒ a + (9 - 1)d = -119

⇒ a + 8d = -119

We know a = -31

⇒ -31 + 8d = -119

⇒ 8d = -119 + 31

⇒ 8d = -88

⇒ d = -11

Now, find a_{32}

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

a_{32} = -31 + (32 - 1) × (-11)

a_{32} = -31 + 31 × (-11)

a_{32} = -31 + ( -341)

a_{32} = -372

Therefore, the value of a_{32} = -372.

## What is the 32nd term of the arithmetic sequence where a_{1} = -31 and a_{9} = -119?

**Summary:**

The 32^{nd} term of the arithmetic sequence where a_{1} = -31 and a_{9} = -119 is a_{32 }= -372.

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