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

Answer:

The 32nd term of arithmetic sequence is -374.

Step-by-step 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.