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
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:
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 :)