What is the 32nd term of the arithmetic sequence where a1 = - 32 and a9 = - 120?

A) - 384

B) - 373

C) - 362

D) - 351

(-120 - (-32)) / (9 - 1) = - 11

a (n) = - 32 - 11 * (n - 1)

a (32) = - 32 - 11 * (32 - 1) = - 373

a = - 32

tn9 = - 120

L = a + (n - 1) * d

-120 = - 32 + (9 - 1) * d Add 32 to both sides.

-88 = 8*d Divide by 8

-11 = d

a_32 = - 32 + (32 - 1) * -11

a_32 = - 32 + 31*-11

a_32 = - 373