Here's the question you clicked on:
Eyad
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 ......
how that c-d is geometric mean of preceding and suceeding term
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 ...
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
I copied it exactly as its written from the book ,though i swear its doesn't make sense ,i totally agreeee
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
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:\[c-b=\frac{a(1-r^{2n})}{1-r}-\frac{a(1-r^{n})}{1-r}=\frac{a(1-r^{2n}-1+r^n)}{1-r}=\frac{a(r^n-r^{2n})}{1-r}=\frac{ar^n(1-r^{n})}{1-r}\]\[\qquad=br^n\]
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
ty ,ima be sure of the 2nd require and then call you back :DDD