Question

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

A) -384
B) -373
C) -362
D) -351

Answer 1

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

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

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

Answer 2

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