Question

Identify the 12th term of a geometric sequence where a1 = 8 and a6 = -8,192.

Answer 1

In a geometric sequence, the nth term, \(\large a_n = a_1 * r^{n-1}\).

\(\large a_1 = 8, n = 6, a_6 = -8192\)

Plug this in and find r.

Then plug n = 12 to find \(\large a_{12}\).

Answer 2

6=8*r^6-1?

Answer 3

\[a_n=a_1*r^{n-1}\]\[a_6=a_1*r^{6-1}\]\[8192=8*r^{5}\]\[r^5=1024\]\[r=\sqrt[5]{1024}\]\[r=4\]Now we just have to replace it\[a_n=a_1*4^{n-1}\]\[a_{12}=a_1*4^{12-1}\]\[a_{12}=8*4^{11}\]The answer is\[\boxed{a_{12}=33554432}\]

Answer 4

thank you :)

Answer 5

You're welcome ;)

Answer 6

can you help with more???? please :)

Answer 7

Sure ;)

Answer 8

Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the nearest hundredth.

Answer 9

a17 ≈ 123,802.31
a17 ≈ 30,707.05
a17 ≈ 19,684.01
a17 ≈ 216,654.05

Answer 10

\[a_n=a_1*r^{n-1}\]\[a_5=a_1*r^{5-1}\]\[150.06=16*r^4\]\[r=\sqrt[4]{9.37875}\]\[r\approx1.75\]now\[a_n=a_1*1.75^{n-1}\]\[a_{17}=a_1*1.75^{17-1}\]\[a_{17}=16*1.75^{16}\]The answer is:\[\boxed{a_{17}=123,802.31}\]

Answer 11

A bacteria culture begins with 13 bacteria which triple in amount every hour. How many bacteria exist in the culture at the 11th hour?
13,312
26,624
767,637
2,302,911

Answer 12

@D3xt3R

Answer 13

\[P(t)=S*3^{t-1}\]\[P(t)=13*3^{t-1}\]If t=11\[P(11)=13*3^{11-1}\]\[P(11)=13*3^{10}\]\[P(11)=13*59,049\]The answer is:\[\boxed{P(11)=767,637}\]

Answer 14

i have two more :) can you help with those? @D3xt3R

Answer 15

Given the functions f(n) = 11 and g(n) = (3/4)n - 1, combine them to create a geometric sequence, an, and solve for the 9th term.
an = (11 • 3.4 )n - 1; a9 ≈ 24.301
an = 11(3/4 )n - 1; a9 ≈ 1.101
an = 11 + (3/4 )n - 1; a9 ≈ 11.100
an = 11 - (3/4 )n - 1; a9 ≈ 9.900

Answer 16

Yes, could you please, give me "Best Response"?! ;)

Answer 17

there you go :)

Answer 18

Just one question is \[g(n)=\left(\frac{3}{4}\right)^{n-1}~~~~or~~~~g(x)=\left(\frac{3}{4}\right)*(n-1)\]?????

Answer 19

g(x)=(3/4)*(n-1)

Answer 20

It's \[g(n)=\left(\frac{3}{4}\right)^{n-1}\]Whatever.\[f(n)*g(n)=11*\left(\frac{3}{4}\right)^{n-1}\]when n=9\[f(9)*g(9)=11*\left(\frac{3}{4}\right)^{9-1}\]The answer is:\[\boxed{f(9)*g(9)=11.100}\]

Answer 21

last one
Given a geometric sequence in the table below, create the explicit formula and list any restrictions to the domain.
n a subn
1 -4
2 20
3 -100

Answer 22

an = -5(-4)n - 1 where n ≥ 1
an = -4(-5)n - 1 where n ≥ 1
an = -4(5)n - 1 where n ≥ -4
an = 5(-4)n - 1 where n ≥ -4

Answer 23

@D3xt3R

Answer 24

\[a_1=-4\]\[a_2=(-4)*(-5)\]\[a_3=(-4)*(-5)*(-5)=(-4)*(-5)^{2}\]Therefore\[a_n=(-4)*(-5)^{n-1}\]Where \[n\geq1\]

Answer 25

thank you :)

Answer 26

You're welcome ;)

Answer 27

A small sign change.
For the very first problem, r = -4 because a6 = -8192
So a12 = -33554432

Answer 28

@D3xt3R @aum what does the r represent in the equation a subn= a1 r^n-1 all divided by 1-r

Answer 29

This formula is the sum of the terms. @ceol1998