What is the 32nd term of the arithmetic sequence where a₁ = -33 and a₉ = -121?

Solution:

The nth term of an arithmetic sequence whose first term is a₁ and common difference is d is given by:

a₁ + (n - 1) d

It is given that

a1 = -33 and a₉ = -121

We know that

a₉ = a₁ + (9 - 1) d

-33 + 8d = - 121

8d = -121 + 33

8d = -88

d = -88/8

d = -11

Now we have to find the 32nd term

a₃₂ = a1 + (32 - 1) d

Substituting the values

a₃₂ = -33 + (31) (-11)

a₃₂ = -33 - 341

So we get

a₃₂ = -374

Therefore, the 32nd term is -374.

What is the 32nd term of the arithmetic sequence where a₁ = -33 and a₉ = -121?

Summary:

The 32nd term of the arithmetic sequence where a₁ = -33 and a₉ = -121 is -374.