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Infinite Geometric Series Help! (1 Viewer)
- Thread starter Zak Ambrose
- Start date
Zak Ambrose
Title Cost $20
ah ok. thanks for the help. much appreciated.
last question then im on to breeze through probability.
the sum of the first 8 terms is 17 times the sum of its first 4 terms. find r.
S8 = 17(S4)=17 [a(1-r^4)/(1-r)]
last question then im on to breeze through probability.
the sum of the first 8 terms is 17 times the sum of its first 4 terms. find r.
S8 = 17(S4)=17 [a(1-r^4)/(1-r)]
Zak Ambrose
Title Cost $20
how did you arrive at
"terms 5-8 are 16 times larger than terms 1-4, and also r^4 times larger"
"terms 5-8 are 16 times larger than terms 1-4, and also r^4 times larger"
Well terms 1-8 is 17 times larger than terms 1-4. So terms [1-8 - 1-4] must be 16 times larger.
If r is the common ratio, then to get from a term to the term n after it you must multiply the number by r^n. So, term 5 is r^4 times greater than term 1 ... term 8 is r^4 times greater than term 4. This means that the sum of terms 5-8 is r^4 times greater than the sum of terms 1-4. (if you take out a factor of r^4 from terms 5-8 you end up with terms 1-4)
If r is the common ratio, then to get from a term to the term n after it you must multiply the number by r^n. So, term 5 is r^4 times greater than term 1 ... term 8 is r^4 times greater than term 4. This means that the sum of terms 5-8 is r^4 times greater than the sum of terms 1-4. (if you take out a factor of r^4 from terms 5-8 you end up with terms 1-4)
Zak Ambrose
Title Cost $20
S8 - S4 = 17S4 - S4
= 16S4
= 16[a(1-r^4)/(1-r)]
am i on the right track?
= 16S4
= 16[a(1-r^4)/(1-r)]
am i on the right track?
the first two lines are right, but I wouldn't bother writing it in the form you wrote it as in the 3rd line.
[S8 - S4] = r^4 S4 = 16S4 is all I would do
* r^1 + r^2 + r^3 + r^4 + r^5 + r^6 + r^7 + r^8
= r^1 + r^2 + r^3 + r^4 + r^4(r^1 + r^2 + r^3 + r^4)
in case you still don't quite understand
[S8 - S4] = r^4 S4 = 16S4 is all I would do
* r^1 + r^2 + r^3 + r^4 + r^5 + r^6 + r^7 + r^8
= r^1 + r^2 + r^3 + r^4 + r^4(r^1 + r^2 + r^3 + r^4)
in case you still don't quite understand
Last edited:
Zak Ambrose
Title Cost $20
dam, im still not getting it.
i've got to...
(r^4)(S4)=16(S4)
r^4 = 16
r = +-2
but what im having trouble with is how i/you got (r^4)(S4)=16(S4)
more precisely - the (r^4)(S4) part.
perhaps if you wrote it out as you would in an exam?
i've got to...
(r^4)(S4)=16(S4)
r^4 = 16
r = +-2
but what im having trouble with is how i/you got (r^4)(S4)=16(S4)
more precisely - the (r^4)(S4) part.
perhaps if you wrote it out as you would in an exam?
Last edited:
=(r^1 + r^2 + r^3 + r^4) + (r^5 + r^6 + r^7 + r^8)
= (r^1 + r^2 + r^3 + r^4) + r^4(r^1 + r^2 + r^3 + r^4)
= (r^4 + 1)S4 = 17S4
"[S8 - S4] = r^4 S4 = 16S4
r^4 = 16, r = +-2" is probably all I would write in an exam
if you divide terms 5-8 by r^4, you get terms 1-4
= (r^1 + r^2 + r^3 + r^4) + r^4(r^1 + r^2 + r^3 + r^4)
= (r^4 + 1)S4 = 17S4
"[S8 - S4] = r^4 S4 = 16S4
r^4 = 16, r = +-2" is probably all I would write in an exam
if you divide terms 5-8 by r^4, you get terms 1-4
Last edited:
Zak Ambrose
Title Cost $20
oh ok. my brain just clicked.
the sum of the terms 5 to 8 = r^4 x S4
I'VE GOT IT!!!
thank you very much for your patience and help.
if you need any help with music or economics gimme a buzz [the only 2 subjects i consider myself worthy of giving internet advice on]
the sum of the terms 5 to 8 = r^4 x S4
I'VE GOT IT!!!
thank you very much for your patience and help.
if you need any help with music or economics gimme a buzz [the only 2 subjects i consider myself worthy of giving internet advice on]