# I need alot of help with geometric sequences and series

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• Mar 27th 2007, 01:24 PM
jonannekeke
I need alot of help with geometric sequences and series
I've answered loads of the questions but I get stuck on the part c and d of them all. so can some one help me asap?

here are the questions.

1) The second term in a geometric series is 80 and the fifth term is 5.12.
r=0.4
a=200
sum to infinity= 333 (1/3)
d)Calculate the difference between the sum to infinity of the series and the sum of the first 14 terms of the series, giving your answer in the form a*10^n, where 1(</=)a<10 and n is an integer.

2)The nth term of a sequence is Un, where Un=95(4/5)^n, n= 1 2 3 ...
c) to 3 significant figures calculate
http://i115.photobucket.com/albums/n...2006/maths.jpg
d) to 3 significant figures calculate the sum to infinity of the series whose 1st term is U1, and whose nth term is Un. ( the dot is next to the 'n')

3)The 3rd and 4th terms of a G.S are 6.4 and 5.12.
r=0.8
a=10
sum to infinity=50
d)Calculate the difference between the sum to infinity and the sum of the first 25 terms of the series.

4)A saving scheme pays 5% per annum compound interest. A deposit of £100 is invested in this scheme at the start of each year. at the start of 3 years= £315.25
b)Find the amount in the scheme at the start of the fortieth year, after the annual deposit has been made.

5)A liquid is kept in a barrel. At the start of the year the barrel is filled with 160 litres of the liquid.Due to evaporation, at the end of every year the amount in the barrel is reduced by 15% of its volume at the start of the year.

At the start of each year a new barrel is filled with 160 litres of liquid so that, at the end of 20 years, there are 20 barrels containing liquid.

c) Calculate the total amount of liquid, to the nearest litre, in the barrels at the end of 20 years.

I know that that is quite a lot of questions but I'm really stuck on them all, I've tried loads of methods but nothing seams to help. All help will be appreciated greatly!

Thanks in advance.
• Mar 27th 2007, 02:34 PM
Jhevon
Quote:

Originally Posted by jonannekeke
I've answered loads of the questions but I get stuck on the part c and d of them all. so can some one help me asap?

here are the questions.

1) The second term in a geometric series is 80 and the fifth term is 5.12.
r=0.4
a=200
sum to infinity= 333 (1/3)
d)Calculate the difference between the sum to infinity of the series and the sum of the first 14 terms of the series, giving your answer in the form a*10^n, where 1(</=)a<10 and n is an integer.

Some things to note:

A geometric series is the sum of the numbers in a geometric progression:
http://upload.wikimedia.org/math/5/f...dc6e9a8d26.png We can find a simpler formula for this sum by multiplying both sides of the above equation by (1 − r), and we'll see that
http://upload.wikimedia.org/math/a/a...94f6faf1fa.png since all the other terms cancel. Rearranging (for http://upload.wikimedia.org/math/0/a...9d228e7585.png) gives the convenient formula for a geometric series:
http://upload.wikimedia.org/math/e/1...542ee8be1a.png Note: If one were to begin the sum not from 0, but from a higher term, say m, then
http://upload.wikimedia.org/math/3/1...74a13148df.png
now to do the problem:

a = 200, r = 0.4

let S14 be the sum of the first 14 terms

using
http://upload.wikimedia.org/math/e/1...542ee8be1a.png

=> S14 = [200(1 - (0.4)^15)]/(1 - 0.4)
= 200(0.999998926)/0.6
= 333.3329753

since the sum to infinity is 333.3333333333333

the difference between the sum to infinity and the sum of the first 4 terms is:

333.3333333 - 333.3329753 = 0.000357967 = 3.58 x 10^-4 or 4 x 10^-4
• Mar 27th 2007, 02:49 PM
Jhevon
Quote:

Originally Posted by jonannekeke

2)The nth term of a sequence is Un, where Un=95(4/5)^n, n= 1 2 3 ...
c) to 3 significant figures calculate
http://i115.photobucket.com/albums/n...2006/maths.jpg
d) to 3 significant figures calculate the sum to infinity of the series whose 1st term is U1, and whose nth term is Un. ( the dot is next to the 'n')

this is a geometric series with
a = 95
r = 4/5

let S15 be the sum of the first 15 terms, then using
http://upload.wikimedia.org/math/e/1...542ee8be1a.png

S15 = [95(1 - (4/5)^16)]/(1 - 4/5)
= (95(0.971852502))/(0.2)
= 461.6299386
= 4.6 x 10^2

(d) the sum to infinity is given by:
S_infinity = a/(1 - r)

= 95/(1 - 4/5)
= 475
= 4.75 x 10^2
• Mar 27th 2007, 02:55 PM
Jhevon
Quote:

Originally Posted by jonannekeke
3)The 3rd and 4th terms of a G.S are 6.4 and 5.12.
r=0.8
a=10
sum to infinity=50
d)Calculate the difference between the sum to infinity and the sum of the first 25 terms of the series.

a = 10, r = 0.8

=> S_infinity = a/(1 - r) = 10/(1 - 0.8) = 50

using S_n = [a(1 - r^(n+1)]/(1 - r)
=> S_25 = [10(1 - 0.8^26)]/(1 - 0.8)
= 49.84888427

So the difference between the sum to infinity and the sum of the first 25 terms is:

S_infinity - S_25 = 50 - 49.84888427 = 0.151115727 = 0.151 or 1.51 x 10^-1
• Mar 27th 2007, 03:17 PM
jonannekeke
Thanks so much, but the answers in my text book don't match with yours for question 2, it says that:
c)367
d)380

Qestion 3 doesn't match either, it says that it should be
D) 0.189 (3 sf)
• Mar 27th 2007, 03:25 PM
Jhevon
Quote:

Originally Posted by jonannekeke
Thanks so much, but the answers in my text book don't match with yours for question 2, it says that:
c)367
d)380

Qestion 3 doesn't match either, it says that it should be
D) 0.189 (3 sf)

so i only got 1 out of three correct, lol. ok, i'll look over them a bit later. i'm kind of busy now
• Mar 27th 2007, 03:26 PM
jonannekeke
thanks for trying though :)
• Mar 27th 2007, 03:37 PM
Jhevon
Quote:

Originally Posted by jonannekeke
thanks for trying though :)

i'm pretty sure the formula's are right though. i guess you can try them and try to work with more accurate decimal approximations than i did
• Mar 27th 2007, 03:41 PM
jonannekeke
I put my little brain into gear and I figured out that a=76 and not 95 in the first one you got wrong because U1=76
and for the other one on the top line of the formula it's just r^n not r^n+1

Thanks for trying though, you did point me in the right direction :)

do you think you willbe able to help with the others? I'm completely puzzled on what to do for them.
• Mar 27th 2007, 03:51 PM
Jhevon
Quote:

Originally Posted by jonannekeke
I put my little brain into gear and I figured out that a=76 and not 95 in the first one you got wrong because U1=76
and for the other one on the top line of the formula it's just r^n not r^n+1

Thanks for trying though, you did point me in the right direction :)

do you think you willbe able to help with the others? I'm completely puzzled on what to do for them.

according to your question, a = 95, oh wait, i see. we started counting at n=1 so our formula should be ar^(n-1), ok, it should be 76 then

but anyway, as long as you got the answer right
• Mar 27th 2007, 03:56 PM
jonannekeke
I dunno, the first part of the question was to find out what U1 was and it was 76, i never really thought I needed to put it oh well.

sorry to bug you, but if you're not too busy could you please try the other two it's getting pretty late (1am-ish) and i need to hand this homework in tomorrow since i haven't been doing any of my homework for weeks lol