Respuesta :

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

Answer:

[tex]a_{32}=-173[/tex]

Step-by-step explanation:

Given : the arithmetic sequence where[tex]a_1 = 13[/tex] and [tex]a_{13} =-59[/tex]

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 [tex]a_n=a+(n-1)d[/tex]

Thus, for the given arithmetic sequence, we have,

First term is [tex]a_1=a= 13[/tex] and

[tex]a_{13}=a+(13-1)d=-59[/tex]

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

[tex]13+(12)d=-59[/tex]

Solving for d, we have,

[tex](12)d=-59-13[/tex]

[tex](12)d=-72[/tex]

Divide by 12 both side, we have,

[tex]d=-6[/tex]

Thus, the common difference is -6.

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

We have,

[tex]a_{32}=13+(32-1)(-6)[/tex]

Simplify, we have,

[tex]a_{32}=13+(31)(-6)[/tex]

[tex]a_{32}=13-186[/tex]

[tex]a_{32}=-173[/tex]