## Eyad 3 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,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 ......

1. experimentX

GS??

Geometric sequence :D

3. experimentX

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

4. experimentX

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 *

7. experimentX

do this first ... then use induction!!

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

9. experimentX

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

10. experimentX

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

11. asnaseer

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

12. asnaseer

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

13. experimentX

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

15. asnaseer

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 ..."

16. asnaseer

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

17. asnaseer

otherwise 3rd term = -(2nd term)

18. asnaseer

which doesn't seem to make sense

19. asnaseer

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

21. asnaseer

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

23. experimentX

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

24. asnaseer

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

maybe ....

26. asnaseer

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 ?

28. asnaseer

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

30. asnaseer

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:$S_{3n}=\frac{a(1-r^{3n})}{1-r}=d$

31. asnaseer

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

34. asnaseer

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

35. asnaseer

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

36. asnaseer

which is a geometric sequence

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

38. asnaseer

yes

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

40. asnaseer

ok - good luck! :)