What is the 32nd term of the arithmetic sequence where a1 = 14 and a13 = −58?

−178
−172
−166
−160

I somehow got -704 can someone help me please.

Respuesta :

a1 = 14
a13 = -58

a13 = a1 + 12d
-58 = 14 + 12d
12d = -58 -14
12d = -72
d = -72/12 = -6

a32 = a1 + 31d = 14 - 31*6 = 14 - 186 = -172

Answer:  The correct option is (B) -172.

Step-by-step explanation:  We are given to find the 32nd term of the arithmetic sequence where the 1st term and 13th term are

[tex]a_1=14,~~~a_{13}=-58.[/tex]

We know that

the n-th term of an arithmetic sequence with first term [tex]a_1[/tex] and common difference [tex]d[/tex] is given by

[tex]a_n=a_1+(n-1)d.[/tex]

According to the given information, we have

[tex]a_1=14,[/tex]

and

[tex]a_{13}=-58\\\\\Rightarrow a_1+(13-1)d=-58\\\\\Rightarrow 14+12d=-58\\\\\Rightarrow 12d=-58-14\\\\\Rightarrow 12d=-72\\\\\Rightarrow d=-6.[/tex]

Therefore, the 32nd term will be

[tex]a_{32}=a_1+(32-1)d=14+31\times(-6)=14-186=-172.[/tex]

Thus, the 32nd term is -172.

Option (B) is correct.