Question

SUPER G.S QUESTION :
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An infinite G.s the sum of the 1st n terms =b ,and the sum of the first 2n terms = c , and the sum of the first 3n terms = d , PROVE THAT (b,c-d,d-c,.....) is an infinite G.s and the sum of any number of its terms cant exceed the sum of the sum of the original sequence up to infinity ......

Answer 1

GS??

Answer 2

Geometric sequence :D

Answer 3

how that c-d is geometric mean of preceding and suceeding term

Answer 4

show that (c-d)^2 = b * (d-c)

Answer 5

@experimentX : he want to prove that these sequence is an infinite G.s ,not a Geometric sequence and since your working on the Geometric Mean that will lead you to the Geometric sequence ...

Answer 6

these sequences are *

Answer 7

do this first ...
then use induction!!

Answer 8

ok ,what about the next require ? i dont even understand what he want !_!

Answer 9

\[ S_n = \frac{n(2a + (n-1)d)}{2}\]
\[ S_{2n} = \frac{2n(2a + (2n-1)d)}{2}\]
\[ S_{3n} = \frac{3n(2a + (3n-1)d)}{2}\]

Find the differences,
\[ S_{2n} - S_{n} = \frac{2n(2a + (2n-1)d)}{2} -\frac{n(2a + (n-1)d)}{2} = \frac{n}{2}(2a + (3n - 1)d) \]
\[ S_{3n} - S_{2n} = \frac{3n(2a + (3n-1)d)}{2} -\frac{2n(2a + (2n-1)d)}{2} = \frac{n}{2}(2a + (5n - 1)d) \]

Answer 10

looks like something went wrong ... verify that these are in Geometric progression first

Answer 11

experimentX: the question says "geometric sequence" not "arithmetic sequence", so the formula for the sum should be:\[S_n=\frac{a(1-r^n)}{1-r}\]

Answer 12

if you use this, then you should be able to do the question.

Answer 13

Oh .. yes that's why i was getting this
http://www.wolframalpha.com/input/?i=255*55+%3D+155^2

Answer 14

yea thats right @asnaseer ,
anyways tysmm of your effort @experimentX

Answer 15

Also, Eyad, I think you may have a mistake in the question. I think it should say:

"PROVE THAT (b, c-b, d-c, .....) is an ..." and not:
"PROVE THAT (b, c-d, d-c, .....) is an ..."

Answer 16

note the 2nd term should be "c-b" not "c-d"

Answer 17

otherwise 3rd term = -(2nd term)

Answer 18

which doesn't seem to make sense

Answer 19

do you agree Eyad?

Answer 20

I copied it exactly as its written from the book ,though i swear its doesn't make sense ,i totally agreeee

Answer 21

must be a misprint

Answer 22

I think i'am gonna leave it and ask da prof. ,and if i got an answer i will surely share it to you guys ... Tysm for your efforts

Answer 23

also something quite not right ... http://www.wolframalpha.com/input/?i=%282^10-1%29%28%282^30-1%29-%282^20-1%29%29+%3D+%282^20-1%29^2

I'm beginning to see things ... i think i should get some sleep .. lol

Answer 24

you should be able to use the formula I gave you to prove this fairly easily

Answer 25

maybe ....

Answer 26

I can do the first step for you if you want?

Answer 27

I dont want to bother you asnaseer i think iam gonna make sure first if the question is written correct then we shall begin again ,what do u think ?

Answer 28

I am 100% sure its a misprint - and it is no bother at all to help you - always a pleasure :)

Answer 29

tysm @asnaseer ,ty too @experimentX ^_^ You guys rocks

Answer 30

ok, we know:\[S_n=\frac{a(1-r^n)}{1-r}=b\]therefore the sum of the first 2n terms must be:\[S_{2n}=\frac{a(1-r^{2n})}{1-r}=c\]similarly, the sum of the first 3n terms must be:\[S_{3n}=\frac{a(1-r^{3n})}{1-r}=d\]

Answer 31

therefore:\[c-b=\frac{a(1-r^{2n})}{1-r}-\frac{a(1-r^{n})}{1-r}=\frac{a(1-r^{2n}-1+r^n)}{1-r}=\frac{a(r^n-r^{2n})}{1-r}=\frac{ar^n(1-r^{n})}{1-r}\]\[\qquad=br^n\]

Answer 32

yea...

Answer 33

yes ,that right i have just proved it ..continue Please

Answer 34

sorry that should be - similarly, you should be able to show:\[d-c=br^{2n}\]

Answer 35

so you will be left with the sequence:\[b,c-b,d-c,...\]which equals:\[b,br^n,br^{2n},...\]

Answer 36

which is a geometric sequence

Answer 37

aha ,now i will be able to prove it can be added to infinite G.s

Answer 38

yes

Answer 39

ty ,ima be sure of the 2nd require and then call you back :DDD

Answer 40

ok - good luck! :)