Question

Find the 30th term in the geometric sequence if t{10}=(512) and t{15}=(16384). HELP!!!

Answer 1

hi there

Answer 2

hello

Answer 3

well it so happens that 512 and 16384 are both powers of 2

Answer 4

2^9 = 512, and 2^14 = 16384, so the sequence appears to be that t{n} = 2^{n-1}

Answer 5

does that help?

Answer 6

ohhh... kind of.

Answer 7

so a = 2?

Answer 8

what formula are you using for the geometric sequence? (i'm not sure what 'a' is in your terminology)

Answer 9

cos I'm following the tn = a*r^n-1

Answer 10

so in this case a is just 1

Answer 11

and r?

Answer 12

2

Answer 13

the nth term is 2 to the power of n-1

Answer 14

so what's the 30th term?

Answer 15

it's gonna be t30 = 1(2)^30-1

Answer 16

right

Answer 17

ohhh..

Answer 18

btw

Answer 19

do I ever use this:

t10 = 512
t15 = 16384

tn = a*r^n-1
t10 = a*r^10-1
512 = a*r^9

Answer 20

tn = a*r^n-1
t15 = a*r^15-1
16 384 = a*r^14

Answer 21

yeah, that's the general way to solve this problem. I was just taking a shortcut since I happened to recognize both of those numbers as being powers of 2, and you need a calculator to take the 14th root of something. since I happened to recognize the numbers it was easy to solve without a calculator.

Answer 22

Divide

16384 ar^14
------ = -------
512 ar^9
32 = r^5
+/- 6.4 = r

Answer 23

ohhh, what if I want to use this method?

Answer 24

sub 6.4 for r in (1):

512 = ar^9
512 = a(6.4)^9
512 = 180a
--- -----
180 180
2. 84 = a

Answer 25

well 16384/512 is 32, and they are 5 terms apart. that means something to the 5th power is equal to 32, i.e. r^5 = 32. The easiest way to solve this then is to take the log of both sides:

5 log(r) = log(32)
log(r) = log(32)/5
r = e^(log(32)/5)
r = 2

Answer 26

sub -6.4 for 4 in (1)

512 = ar^9
512 = a(-6.4)^9
512 = -180a
--- -----
-180 -180

- 2. 84 = a

Answer 27

Since a = 2.84 & r = 6.4, the first three terms are: 2. 84, 18.18, 116. 33, ...
Since a = -2.84 & r = -6.4 the first three terms are: -2.84, 18.18, -116. 33, ...

Answer 28

we appear to be working on different problems

Answer 29

really?

Answer 30

You wrote:
32=r^5
r=6.4

This is wrong, how did you get that value of r?

Answer 31

ohhh.. how do I do that?

Answer 32

Oh wait you just divided. That's not right. r to the 5th power is 32, not r times 5.

Answer 33

Do you know how to use logarithms?

Answer 34

nope.

Answer 35

are you allowed to use a calculator?

Answer 36

yes, but do you know how to do it by paper-&-pencil?

Answer 37

I'd like how to know both methods.

Answer 38

There's no easy way to do it with pencil because you have to take the 5th root of 32, which you need to do with logarithms

Answer 39

ohh, okay. so by using the calculator?

Answer 40

take the 5th root of 32

Answer 41

huh?

Answer 42

how?

Answer 43

hello?