Question

What is the 32nd term of the arithmetic sequence where a1 = −32 and a9 = −120?

Answer 1

-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

Answer 2

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`