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SUPER G.S QUESTION : ___________________ 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 ......

Mathematics
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GS??
Geometric sequence :D
how that c-d is geometric mean of preceding and suceeding term

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Other answers:

show that (c-d)^2 = b * (d-c)
@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 ...
these sequences are *
do this first ... then use induction!!
ok ,what about the next require ? i dont even understand what he want !_!
\[ S_n = \frac{n(2a + (n-1)d)}{2}\] \[ S_{2n} = \frac{2n(2a + (2n-1)d)}{2}\] \[ S_{3n} = \frac{3n(2a + (3n-1)d)}{2}\] Find the differences, \[ S_{2n} - S_{n} = \frac{2n(2a + (2n-1)d)}{2} -\frac{n(2a + (n-1)d)}{2} = \frac{n}{2}(2a + (3n - 1)d) \] \[ S_{3n} - S_{2n} = \frac{3n(2a + (3n-1)d)}{2} -\frac{2n(2a + (2n-1)d)}{2} = \frac{n}{2}(2a + (5n - 1)d) \]
looks like something went wrong ... verify that these are in Geometric progression first
experimentX: the question says "geometric sequence" not "arithmetic sequence", so the formula for the sum should be:\[S_n=\frac{a(1-r^n)}{1-r}\]
if you use this, then you should be able to do the question.
Oh .. yes that's why i was getting this http://www.wolframalpha.com/input/?i=255*55+%3D+155^2
yea thats right @asnaseer , anyways tysmm of your effort @experimentX
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 ..."
note the 2nd term should be "c-b" not "c-d"
otherwise 3rd term = -(2nd term)
which doesn't seem to make sense
do you agree Eyad?
I copied it exactly as its written from the book ,though i swear its doesn't make sense ,i totally agreeee
must be a misprint
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
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
you should be able to use the formula I gave you to prove this fairly easily
maybe ....
I can do the first step for you if you want?
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 ?
I am 100% sure its a misprint - and it is no bother at all to help you - always a pleasure :)
tysm @asnaseer ,ty too @experimentX ^_^ You guys rocks
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:\[S_{3n}=\frac{a(1-r^{3n})}{1-r}=d\]
therefore:\[c-b=\frac{a(1-r^{2n})}{1-r}-\frac{a(1-r^{n})}{1-r}=\frac{a(1-r^{2n}-1+r^n)}{1-r}=\frac{a(r^n-r^{2n})}{1-r}=\frac{ar^n(1-r^{n})}{1-r}\]\[\qquad=br^n\]
yea...
yes ,that right i have just proved it ..continue Please
sorry that should be - similarly, you should be able to show:\[d-c=br^{2n}\]
so you will be left with the sequence:\[b,c-b,d-c,...\]which equals:\[b,br^n,br^{2n},...\]
which is a geometric sequence
aha ,now i will be able to prove it can be added to infinite G.s
yes
ty ,ima be sure of the 2nd require and then call you back :DDD
ok - good luck! :)

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