Question

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

Answer 1

Firstly we have to find d:
$d = \frac{a_{n}-a_{m}}{n-m}$  ⇒
$d = \frac{-120-(-32)}{9-1} = \frac{-88}{8} = -11$
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 $a_{32} = -373$

Answer 2

Answer:

-373

Step-by-step 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