Question

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

Answer 1

`a_1=-33;\ a_9=-121\\\\a_9-a_1=8d\\\\8d=-121-(-33)\\8d=-121+33\\8d=-88\ \ \ \ |divide\ both\ sides\ by\ 8\\d=-11\\\\a_(32)=a_1+31d\\\\a_(32)=-33+31\cdot(-11)=-33-341=-374\\\\Answer:\boxed{a_(32)=-374}`

Answer 2

The 32nd term of arithmetic sequence is -374.

Explanation:

Given : the arithmetic sequence where
`a_1 = -33` and
`a_(9) =-121`

We have to find the 32nd term of the arithmetic sequence.

For the arithmetic sequence having first term 'a' and common difference 'd' the general term is defined by
`a_n=a+(n-1)d`

Thus, for the given arithmetic sequence, we have,

First term is -33

`a_(9)=a+(9-1)d=-121`

We can calculate the common difference by putting a = -33 in above, we have,

-33 + 8 d = -121

Solving for d, we have,

8d = -121 + 33

⇒ 8d = -88

⇒ d = - 11

Thus, the common difference is -11.

For 32nd term, Put a = -33 , d = -11 and n = 32 in
`a_n=a+(n-1)d`

We have,

`a_(32)=-33+(32-1)(-11)`

Simplify, we have,

`a_(32)=-33+(31)(-11)`

`a_(32)=-33-341=-374`

`a_(32)=-374`

Thus, the 32nd term of arithmetic sequence is -374.