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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,cd,dc,.....) 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 ......
 one year ago
 one year ago
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,cd,dc,.....) 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 ......
 one year ago
 one year ago

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experimentXBest ResponseYou've already chosen the best response.0
how that cd is geometric mean of preceding and suceeding term
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
show that (cd)^2 = b * (dc)
 one year ago

EyadBest ResponseYou've already chosen the best response.0
@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 ...
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
do this first ... then use induction!!
 one year ago

EyadBest ResponseYou've already chosen the best response.0
ok ,what about the next require ? i dont even understand what he want !_!
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ S_n = \frac{n(2a + (n1)d)}{2}\] \[ S_{2n} = \frac{2n(2a + (2n1)d)}{2}\] \[ S_{3n} = \frac{3n(2a + (3n1)d)}{2}\] Find the differences, \[ S_{2n}  S_{n} = \frac{2n(2a + (2n1)d)}{2} \frac{n(2a + (n1)d)}{2} = \frac{n}{2}(2a + (3n  1)d) \] \[ S_{3n}  S_{2n} = \frac{3n(2a + (3n1)d)}{2} \frac{2n(2a + (2n1)d)}{2} = \frac{n}{2}(2a + (5n  1)d) \]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
looks like something went wrong ... verify that these are in Geometric progression first
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
experimentX: the question says "geometric sequence" not "arithmetic sequence", so the formula for the sum should be:\[S_n=\frac{a(1r^n)}{1r}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
if you use this, then you should be able to do the question.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
Oh .. yes that's why i was getting this http://www.wolframalpha.com/input/?i=255*55+%3D+155^2
 one year ago

EyadBest ResponseYou've already chosen the best response.0
yea thats right @asnaseer , anyways tysmm of your effort @experimentX
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
Also, Eyad, I think you may have a mistake in the question. I think it should say: "PROVE THAT (b, cb, dc, .....) is an ..." and not: "PROVE THAT (b, cd, dc, .....) is an ..."
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
note the 2nd term should be "cb" not "cd"
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
otherwise 3rd term = (2nd term)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
which doesn't seem to make sense
 one year ago

EyadBest ResponseYou've already chosen the best response.0
I copied it exactly as its written from the book ,though i swear its doesn't make sense ,i totally agreeee
 one year ago

EyadBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
also something quite not right ... http://www.wolframalpha.com/input/?i=%282^101%29%28%282^301%29%282^201%29%29+%3D+%282^201%29^2 I'm beginning to see things ... i think i should get some sleep .. lol
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
you should be able to use the formula I gave you to prove this fairly easily
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
I can do the first step for you if you want?
 one year ago

EyadBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
I am 100% sure its a misprint  and it is no bother at all to help you  always a pleasure :)
 one year ago

EyadBest ResponseYou've already chosen the best response.0
tysm @asnaseer ,ty too @experimentX ^_^ You guys rocks
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
ok, we know:\[S_n=\frac{a(1r^n)}{1r}=b\]therefore the sum of the first 2n terms must be:\[S_{2n}=\frac{a(1r^{2n})}{1r}=c\]similarly, the sum of the first 3n terms must be:\[S_{3n}=\frac{a(1r^{3n})}{1r}=d\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
therefore:\[cb=\frac{a(1r^{2n})}{1r}\frac{a(1r^{n})}{1r}=\frac{a(1r^{2n}1+r^n)}{1r}=\frac{a(r^nr^{2n})}{1r}=\frac{ar^n(1r^{n})}{1r}\]\[\qquad=br^n\]
 one year ago

EyadBest ResponseYou've already chosen the best response.0
yes ,that right i have just proved it ..continue Please
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
sorry that should be  similarly, you should be able to show:\[dc=br^{2n}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
so you will be left with the sequence:\[b,cb,dc,...\]which equals:\[b,br^n,br^{2n},...\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
which is a geometric sequence
 one year ago

EyadBest ResponseYou've already chosen the best response.0
aha ,now i will be able to prove it can be added to infinite G.s
 one year ago

EyadBest ResponseYou've already chosen the best response.0
ty ,ima be sure of the 2nd require and then call you back :DDD
 one year ago
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