# What is the 32^{nd} term of the arithmetic sequence where a_{1} = 12 and a_{13} = -60?

**Solution:**

The n^{th} term of an arithmetic sequence whose first term is a_{1} and common difference is d is

⇒ a_{1} + (n - 1)d

It is given that

a_{1} = 12 and a_{13} = -60

We know that

a_{13} = a_{1} + (13 - 1)d

12 + 12d = -60

12d = -60 - 12

12d = -72

d = -72/12

d = -6

Now we have to find the 32^{nd} term

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

Substituting the values

a_{32} = 12 + (31)(-6)

a_{32} = 12 - 186

So we get,

a_{32} = -174

Therefore, the 32^{nd} term is -174.

## What is the 32^{nd} term of the arithmetic sequence where a_{1} = 12 and a_{13} = -60?

**Summary:**

The 32^{nd} term of the arithmetic sequence where a_{1 }= 12 and a_{13} = -60 is -174.