Question

What is the 32nd term of the arithmetic sequence where a1 = 13 and a13 = –59?

Answer 1

`a_(32)=-173`

Explanation:

Given : the arithmetic sequence where
`a_1 = 13` and
`a_(13) =-59`

We have to find the 32nd term of the arithmetic sequence.

Since, the general arithmetic sequence having first term 'a' and common difference 'd' is given by
`a_n=a+(n-1)d`

Thus, for the given arithmetic sequence, we have,

First term is
`a_1=a= 13` and

`a_(13)=a+(13-1)d=-59`

Calculate the common difference by putting a = 13 in above, we have,

`13+(12)d=-59`

Solving for d, we have,

`(12)d=-59-13`

`(12)d=-72`

Divide by 12 both side, we have,

`d=-6`

Thus, the common difference is -6.

For 32nd term, Put a = 13 , d = -6 and n = 32 in
`a_n=a+(n-1)d`

We have,

`a_(32)=13+(32-1)(-6)`

Simplify, we have,

`a_(32)=13+(31)(-6)`

`a_(32)=13-186`

`a_(32)=-173`

Answer 2

we are given with the data in an arithmetic sequence of a1 = 13 and a13 = –59 and ought to find the 32nd term of the sequence

an = a1 + (n-1) *d
-59 = 13 + (12)*d
d = -6

an = a1 + (n-1) *d
a32 = 13 + 31* -6
a32= -173