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 2 years ago
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 ......
 2 years 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 ......

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experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0how that cd is geometric mean of preceding and suceeding term

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0show that (cd)^2 = b * (dc)

Eyad
 2 years ago
Best 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 ...

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0do this first ... then use induction!!

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0ok ,what about the next require ? i dont even understand what he want !_!

experimentX
 2 years ago
Best 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, \[ S_{2n}  S_{n} = \frac{2n(2a + (2n1)d)}{2} \frac{n(2a + (n1)d)}{2} = \frac{n}{2}(2a + (3n  1)d) \] \[ S_{3n}  S_{2n} = \frac{3n(2a + (3n1)d)}{2} \frac{2n(2a + (2n1)d)}{2} = \frac{n}{2}(2a + (5n  1)d) \]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0looks like something went wrong ... verify that these are in Geometric progression first

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2if you use this, then you should be able to do the question.

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

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0yea thats right @asnaseer , anyways tysmm of your effort @experimentX

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2Also, 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 ..."

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2note the 2nd term should be "cb" not "cd"

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2otherwise 3rd term = (2nd term)

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2which doesn't seem to make sense

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

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0also 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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2I can do the first step for you if you want?

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0I 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 ?

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

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0tysm @asnaseer ,ty too @experimentX ^_^ You guys rocks

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2ok, 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:\[S_{3n}=\frac{a(1r^{3n})}{1r}=d\]

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

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0yes ,that right i have just proved it ..continue Please

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

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

asnaseer
 2 years ago
Best ResponseYou've already chosen the best response.2which is a geometric sequence

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0aha ,now i will be able to prove it can be added to infinite G.s

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