Find the 5th term of the geometric sequence 2,8,32,...?
Thank you for your help!
2007-04-18 15:32:14 · 4 answers · asked by Mrs.Sizemore 2 in Science & Mathematics ➔ Mathematics
2, 8, 32
Each term is 4 time the previous term. So the sequence continues
4*32 = 128
4*128 = 512
The fifth term is 512.
2007-04-18 15:54:48 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
an = a1 * r^(n - 1)
an = 2 * r^(2 - 1)
8 = 2 * r
r = 4
an = 2 * r^(3 - 1)
32 = 2 * r^2
r^2 = 16
r = 4
The formula is
an = 2 * 4^(n - 1)
or
an = 2 * (2^2)^(n - 1)
an = (2^1)(2^(2(n - 1)))
an = (2^1)(2^(2n - 2))
an = 2^(1 + (2n - 2))
an = 2^(2n - 1)
a5 = 2^(2(5) - 1)
a5 = 2^(10 - 1)
a5 = 2^9
a5 = 512
2007-04-18 23:07:39 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
2^1, 2^3, 2^5, 2^7, 2^9
2^9 = 512
2007-04-18 22:35:54 · answer #3 · answered by zeb 4 · 0⤊ 0⤋
That is no calculus.
Determine the common ratio: it is 4.
You can then iterate to find the 5th term:
2,8,32,128, 512.
512 is your answer.
2007-04-18 22:37:00 · answer #4 · answered by Joshua B 2 · 0⤊ 0⤋