Question

What is the 32nd term of the arithmetic sequence where a1 = 12 and a13 = -60

Answer 1

Answer:

- 174

Step-by-step explanation:

The nth term of an arithmetic  sequence is given as

Tn = a + (n - 1)d

where Tn is the nth term

a is the first term , n is the number of term and d is the common difference. As such,

a13 = a + (13 - 1)d

= a + 12d

Given that a1 = 12 and a13 = -60

-60 = 12 + 12d

12d = -72

d = -6

Hence a32 which is the 32nd term

= 12 + (32 - 1)-6

= 12 + (-186)

= - 174

Answer 2

Answer:

The 32nd term of Arithmetic sequence is -174

Step-by-step explanation:

Given: $a_1=12, a_{13}=-60$

We are given two term of the Arithmetic sequence.

Formula:

$a_n=a+(n-1)d$

For  $a_1=12$

$a=12$

For $a_{13}=-60$

$a+12d=-60$

Using two equation solve for a and d

$12+12d=-60$

$1+d=-5$

$d=-6$

We need to find 32nd term

$a_{32}=a+31d$

$a_{32}=12+31(-6)$

$a_{32}=12-186$

$a_{32}=-174$

Hence, The 32nd term of Arithmetic sequence is -174