The geometric sequence will be greater than the arithmetic sequence at the 8th term
How to determine when the geometric sequence will be greater than the arithmetic sequence?
Geometric sequence
Here, we have
First term, a = 3
Common ratio, r = 3 i.e. 9/3 = 3
So, the n-th term of the geometric sequence is
G(n) = 3 * 3^n-1
Arithmetic sequence
Here, we have
First term, a = 300
Common difference, d = 500 i.e. 800 - 300 = 500
So, the n-th term of the arithmetic sequence is
A(n) = 300 + 500(n - 1)
When G(n) is greater than A(n), we have:
3 * 3^n-1 > 300 + 500(n - 1)
This gives
3^n > 300 + 500(n - 1)
Using a graphing calculator, we have:
n > 7.4
The next integer value of n is
n = 8
Hence, the geometric sequence will be greater than the arithmetic sequence at the 8th term
Read more about sequence at:
https://brainly.com/question/7882626
#SPJ1
