Question

What is the 32nd term of the arithmetic sequence where a1 = -32 and a9 = -120? Choose one answer. A) -384. B) -373. C) -362. D) -351

Answer 1

a 1 = - 32,  a 9 = - 120
a 9 = a 1 + 8 d
- 120 = - 32 + 8 d
8 d = - 88
d = - 88 : 8
d = - 11
a 32  = a 1 + 31 d = - 32 + 31 * ( -11 ) = - 32 - 341 = - 373
Answer: B ) - 373 

Answer 2

Answer:

Step-by-step explanation:

Alright, lets get started.

The formula for nth term of arithmetic sequence is :

$t_{n}=a+(n-1)d$

First term is -32

Ninth term is :

$t_{9}=-32 + (9-1)*d$

$-120=-32+8*d$

$8d=-120+32$

$8d=-88$

$d=-11$

32th term is :

$t_{32}=-32 + (32-1)*(-11)$

$t_{32}=-32-341$

$t_{32}=-373$

So the answer is -373   :    Answer

Hope it will help :)