Answer:
yes it is the 10th term in the series
Step-by-step explanation:
The nth term of a geometric sequence is
[tex]a_{n}[/tex] = a₁ [tex](r)^{n-1}[/tex]
where a₁ is the first term and r the common ratio
Here a₁ = 7 and r = [tex]\frac{a_{2} }{a_{1} }[/tex] = [tex]\frac{14}{7}[/tex] = 2 , then
[tex]a_{n}[/tex] = 7 [tex](2)^{n-1}[/tex]
Equate [tex]a_{n}[/tex] to 3584 and solve for n
7 [tex](2)^{n-1}[/tex] = 3584 ( divide both sides by 7 )
[tex]2^{n-1}[/tex] = 512 , that is
[tex]2^{n-1}[/tex] = [tex]2^{9}[/tex]
Since the bases on both sides are equal, both 2 , then equate the exponents
n - 1 = 9 ( add 1 to both sides )
n = 10
3584 is the 10th term in the series
