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
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.