A. 1/128
B. 1/256
C. 1/384
D. 1/768
2006-08-16 03:43:59 · 9 answers · asked by shahad 1 in Science & Mathematics ➔ Mathematics
C. 1/384
1/3, 1/6, 1/12, 1/24, 1/48, 1/96, 1/192, 1/384
2006-08-16 03:49:36 · answer #1 · answered by q_midori 4 · 1⤊ 0⤋
Geometric series takes the variety an = a r^n. subsequently a3 = a r^3, a8 = a r^8, so a8/a3 = (a r^8)/(a r^3) = r^5. on your question, a8/a3 = 8748/36 = 243, so r^5 = 243, and r = 3. Then from a3 = a three^3 = 36, we get a = 36/27 = 4/3. subsequently series is an = (4/3) 3^n = 4 (3)^(n-a million) and now finding a1, a10 is trivial. it is, a1 = 4, and a10 = 4(3^9)=78732.
2016-11-04 22:40:25 · answer #2 · answered by Anonymous · 0⤊ 0⤋
C 1/384
2006-08-16 03:55:43 · answer #3 · answered by Marc B 3 · 0⤊ 0⤋
Here first term =a=1/3, second term = 1/6
common ratio = r =1/2 which is got dividing second term by first term
using the formula
nth term = first term multiplied by (n-1)th power of common ratio
8th term =( 1/3)multiplied by (1/2)^7
=(1/3) (1/128)
=1/384
2006-08-16 03:59:31 · answer #4 · answered by Amar Soni 7 · 0⤊ 0⤋
well its called geometric progression. there is a formula for that.
(i wish there is microsoft equation editor here)
nth term = a r ^ (n-1)
a is the first term
r is the ratio (in this case, (1/6)/(1/3)
n is 8 (a8, 8th term of the sequence)
so...
nth term = a r ^ (n-1) (a times r raised to n minus 1)
=(1/3) (1/2)^ (8-1)
= 1/384
hope i helped you
2006-08-16 04:03:28 · answer #5 · answered by harry 2 · 0⤊ 0⤋
1/384
2006-08-16 03:51:03 · answer #6 · answered by Fredrick Carley 2 · 0⤊ 0⤋
an = a1 * r^(n - 1)
a8 = (1/3) * (1/2)^(8 - 1)
a8 = (1/3) * (1/2)^7
a8 = (1/384)
ANS : C.
2006-08-16 08:42:54 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
This is too tough 4 me. I'm only 14 and I don't even understand Geometrical Construction.
2006-08-16 03:49:22 · answer #8 · answered by ErC 4 · 0⤊ 0⤋
i'm gonna say answer c
2006-08-16 03:50:25 · answer #9 · answered by mikedog_mikedog 2 · 0⤊ 0⤋