Question

What are the explicit equation and domain for a geometric sequence with a first term of 4 and a second term of −12? an = 4(−3)n − 1; all integers where n ≥ 1 an = 4(−3)n − 1; all integers where n ≥ 0 an = 4(36)n − 1; all integers where n ≥ 1 an = 4(36)n − 1; all integers where n ≥ 0

Answer 1

Answer:

$a_{n} = 4 \times (-3)^{n-1}$; all integers where n ≥ 1

Step-by-step explanation:

We are given that the first term of the sequence is 4 i.e. $a_{1} =4$ and second term is -12 i.e. $a_{2} = -12$.

Now, the nth term of geometric sequence is given by,

$a_{n} = a_{n-1} \times r$, where r is the common ratio of the sequence.

So, using the given values we get,

$a_{2} = a_{1} \times r$

i.e. $r = \frac{a_{2} }{a_{1} }$

i.e. $r = \frac{-12}{4}$

i.e. r = -3.

Now the explicit formula for the geometric equation is given by,

$a_{n} = a_{1} \times r^{n-1}$.

i.e. $a_{n} = 4 \times(-3)^{n-1}$, where $n\geq 1$

Hence option first is correct.

Answer 2

The geometric sequence is given by:
an=ar^(n-1)
where:
a=first term
r=common ratio
n is the nth term
given that a=4, and second term is -12, then
r=-12/4=-3
hence the formula for this case will be:
an=4(-3)^(n-1)
where n≥1