Find the eighth term of the sequence whose first three terms are 4,2,1.?

Answers

Answer #1 | 17/04 2017 19:11

I'm assuming that is a geometric sequence where each term is 1/2 of the prior term.
With only 8 terms, you could do this manually.
4, 2, 1, 1/2, 1/4, 1/8, 1/16, 1/32
Or you could use the general formula:
a[n] = ar^(n-1)
a : first term (4)
r : common ratio (½)
a[8] = 4*½^7
= 4 * 1/128
= 1/32

Positive: 50 %

Answer #2 | 17/04 2017 19:18

1__2__3_____4_____5____6____7_____8
4__2__1____1/2____1/4__1/8__1/16___1/32
8th term is 1/32

Positive: 0 %

Answer #3 | 17/04 2017 19:18

4, 2, 1, 1/2, ...
an = 2^(3 - n)
a8 = 1/32

Positive: 0 %

Answer #4 | 17/04 2017 19:12

#1 = a = 4
#2 = ar = 2
#3 = ar^(2) = 1
Hence
#3 / #2 = ar^2/ar = 1/2 = r ( By cancelling down).
#n = ar^(n-1)
#8 = ar^7 = 4 x ( 1/2)^7 = 4/128 = 1/32

Answer #1| 17/04 2017 19:1150 %