75.5k views
5 votes
What is the 32nd term of the arithmetic sequence where a1 = −32 and a9 = −120?

2 Answers

3 votes

Answer:

-373

Explanation:

A

n

=

A

1

+

d

(

n

1

)

A

1

=

32

A

9

=

120

A

9

=

32

+

d

(

9

1

)

120

=

32

+

d

(

8

)

88

=

8

d

d

=

11

A

32

=

32

+

(

11

)

(

32

1

)

A

32

=

32

+

(

11

)

(

31

)

A

32

=

32

+

341

A

32

=

373

User Florian Klein
by
8.6k points
3 votes
firstly we have to find d:

d = (a_(n)-a_(m))/(n-m)

d = (-120-(-32))/(9-1) = (-88)/(8) = -11
now, when we know d, we can find the 32nd term by using a following formula:

a_(n)=a_(1)+(n-1)*d

a_(32) =-32+31*(-11) = -32-341 = -373
the answer is
a_(32) = -373
User Preetie
by
8.0k points

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