What is the 32nd term of the arithmetic sequence where a1 = −34 and a9 = −122? (1 point)

Answer:The 32nd term of the arithmetic sequence is -386.Step-by-step explanation:Given:   The arithmetic sequence where [tex]a_1=-34[/tex] and  [tex]a_9=-122[/tex] We have to find the 32nd term of the arithmetic sequence.Consider the given sequence with [tex]a_1=-34[/tex] and [tex]a_9=-122[/tex] We know , For a given sequence in an Arithmetic sequence with first term [tex]a_1[/tex] and common difference d , we have, [tex]a_n=a_1+(n-1)d[/tex]We first find the common difference "d".[tex]a_9=-122[/tex] [tex]a_9=a_1+(9-1)d[/tex] [tex]a_1=-34[/tex] , we have,[tex]-122=-34+8d[/tex] Solve for d , we have,-88= 8dd = - 11Thus, 32nd term is [tex]a_{32}=a_1+(32-1)d[/tex][tex]a_{32}=-34+32\cdot (-11)[/tex][tex]a_{32}=-386[/tex]Thus, The 32nd term of the arithmetic sequence is -386.