Question

Could somebody possibly help me with Geometric series ? :)

Answer 1

Sure, whats your question ? :)

Answer 2

There was this chart we had for algebra. I went over it a few times trying to get certain answers , yet i still don't know how to do it.

Answer 3

Post your exact question or screenshot of it :)

Answer 4

Well here's the chart. :)
Ball 1 Description Ball 2 Description Ball 3 Description
Height 1
(starting point)
3 ft 3ft 3ft


Height 2
2.4 1.9 0.9


Height 3
1.9 1.2 0.3

Height 4
1.5 0.1 0.8

Height 5
1.2 0.5 0.02


The first question would be, What is the average common ratio between the successive height values of ball 1? Ball 2? Ball 3? Experimental errors may cause common ratios to have some variances within the data for one ball. Use the average common ratio

Answer 5

you know what is the meaning of common ratio ?

Answer 6

The common ratio is the ratio of a term to the previous term. This ratio is usually indicated by the variable r. :)

Answer 7

Common ratio is the ratio between consecutive terms.
like example : 1,4,16,64....
here the common ratio is r = 4/1 = 16/4 = 64/16 = 4

r = common ratio = 2nd term / 1st term = 3rd term / 2nd term = 4th term/3rd term = ...and so on

see if this makes sense ?

Answer 8

I'm sort of getting the gist of this. So, they're asking for the common ratios of the heights of the balls. So the numbers they have would be like - 3/3, 0.9/1.9, 0.3/1.2 and so on ?

Answer 9

just concentrate on ball1 first

from the table, for ball1 you have
3,2.4,1.9,1.5,1.2

what is ratio r between 2nd term and 1st term ??
what is ratio r between 3rd term and 2nd term ??
what is ratio r between 4th term and 3rd term ??

Answer 10

and lastly
what is ratio r between 5th term and 4th term ??

Answer 11

For ball 1, the ratios between the 2nd and 1st term would be, 0.8
The 3rd and 2nd would be, 0.79
and the 4th and 3rd term would be 0.78
and the 5th and 4th term would be 0.8 again. right ?

Answer 12

1.5/1.9 is approximately 0.79 only.
and ues, all others are correct!
so, what is their approximate average ?

Answer 13

*and yes**

Answer 14

That would be 0.78 is the average. :)

Answer 15

so, i guess you would agree that the average common ratio for ball1 is approximately = 0.8

do the same thing for ball2 and ball 3 :)

Answer 16

0.8,0.79,0.79,0.8 ...average is 0.795
take it as 0.8 only :)

Answer 17

Aw, i was close . :)
Could you also possible help me with two more questions if you can ? :)

Answer 18

sure, but i had one doubt, can i ask ?

Answer 19

Of course. :)

Answer 20

if you are that other girl, who is the first girl ??? :P

Answer 21

lol, just kidding, ask your questions :)

Answer 22

oh, i see what you did there. cx
and the next questions have to do with the same graph. :)
#2 - What is the height of each ball on the fifth bounce (i.e., Height 6)? Use the geometric sequence formula, an = a1r^n – 1 and show your work.

Answer 23

\(\Large a_n = a_1r^{n – 1 }\)

a1 = 1st term = 3
r = 0.8 (we calculated)
n= number of term = 6 (as we need height 6)

just plug in values!

Answer 24

So it would be an = 3(0.8)^5 ?

Answer 25

thats correct
calculate it

Answer 26

That would be 80. right ? :)

Answer 27

80 ? :O
\(\Large 3 \times 0.8^5 =0.983\)

Answer 28

ohohohoh. i see i did that all wrong :0

Answer 29

Now that i did that the correct way, i see i got the same answer.

Answer 30

good! :)
have anymore questions ?

Answer 31

One last one if you don't mind. :)

Answer 32

sure ask

Answer 33

#3 - What is the total distance of the height each ball has traveled in the first five heights? Use the geometric series formula, Sn = \[a_{1}- a_{1}r^n \over 1 - r\]

Answer 34

you already know all the variables
just plug them in!


a1 = 1st term = 3
r = 0.8 (we calculated)
n= number of term = 5 (as they want sum of first 5 heights)

Answer 35

I see that now, i just needed to see if i was correct. Thank you so much for all your help. :)

Answer 36

Welcome ^_^

and bdw, Vivian, since you're new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

if you have any doubts browsing this site, you can ask me :)

Answer 37

Thank you so much. Will do :)