Question

Identify the 16th term of a geometric sequence where a1 = 4 and a8 = −8,748. (2 points)

a
−172,186,884

b
−57,395,628

c
57,395,628

d
172,186,884

Answer 1

a1=a=4
a8=a+7d=8748 ----1
substitute a in 1
4+7d=8748
7d=8748-4
7d=8744
d=8744÷7

Answer 2

Option b - the 16th term of GP is −57,395,628.

Step-by-step explanation:

Given : A geometric sequence where $a_1 = 4$ and $a_8 = -8748$.

To find : Identify the 16th term of a geometric sequence ?

Solution :

The 8th term of GP is $a_8=ar^7$.

Substitute, $a_1 = 4$ and $a_8 = -8748$.

$-8748=(4)(r)^7$

$r^7=-2187$

$r^7=(-3)^7$

On comparing,

$r=-3$

The common ratio is r=-3.

The 16th term of GP is $a_{16}=ar^{15}$

$a_{16}=(4)(-3)^{15}$

$a_{16}=4\times -14348907$

$a_{16}=-57395628$

Therefore, the 16th term of GP is −57,395,628.

So, option b is correct.

#Learn more

Determine the n-th term of gp if the 4-th term is 24 and 7-th term is 192

https://brainly.in/question/14942373