Need help on another Maths question Core 2 this time

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Going through an OCR June 2007 core 2 past papaer

On Q7.

a) In an arithmetic progression, the first term is 12 and the sum of the first 70 terms is 12915. Find the common difference/

12=a 12915= total 70 term

12915= 70/2(24+(70-1)d)

12915=35(24+69d)
12915=840+2415d
12075=2415d
d=5

b) In a geometric progression, the second term is -4 and the sum infinity is 9. Find the Common ratio
~<<<<<infinity

ar= -4

then i know i have to put it into the equation s~=a/1-r

9=-4/1-r I am guessing you some how get two sets of brackets that will give you value of r

I don't really know where to go from here so if someone could explain the steps to me then i would be very grateful

Thanks
 
You are correct that ar=-4
After that you seem to have gone wrong?

Sum to infinity is indeed a/1-r
Mess about with ar=-4 to give you a = -4/r
so you get 9 = (-4/r)/(1-r)
multiply by 1-r
9-9r=-4/r
multiply by r
9r-9r^2=-4
9r^2-9r-4=0
factorize.

Not a very nice factorization. I may well have done something wrong here! Got core 2 AQA tommorow and sequences/series is my weekpoint. Am pretty good at everything else!
 
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Urgh, progressions.

Top tip: Do not spend the whole of AS Maths chatting up the good looking blonde girl..... I did and last summer got a U in C1 and a U in C2 :D

Retook in Jan this year and got A & B.... and progressions didn't come up :cool:
 
Progressions are the one bit I seriously can't do. I just can't get my head around them. Completely ace all other sections. Hope they don't come up majorly tommorrow :(
 
You are correct that ar=-4
After that you seem to have gone wrong?

Sum to infinity is indeed a/1-r
Mess about with ar=-4 to give you a = -4/r
so you get 9 = (-4/r)/(1-r)
multiply by 1-r
9-9r=-4/r
multiply by r
9r-9r^2=-4
9r^2-9r-4=0
factorize.

Not a very nice factorization. I may well have done something wrong here! Got core 2 AQA tommorow and sequences/series is my weekpoint. Am pretty good at everything else!


Thanks for that got the answer. I went wrong when got ar=-4 didnt make a subject. In the end you get R= -1/3 or 4/4
r=-1/3

By the looks of things i think you will do pretty well good luck mate.


I have a B in C1 retake yesterday so hopefully it pulls it up to an A but the last question was horrendous
 
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You are correct that ar=-4
After that you seem to have gone wrong?

Sum to infinity is indeed a/1-r
Mess about with ar=-4 to give you a = -4/r
so you get 9 = (-4/r)/(1-r)
multiply by 1-r
9-9r=-4/r
multiply by r
9r-9r^2=-4
9r^2-9r-4=0
factorize.

Not a very nice factorization. I may well have done something wrong here! Got core 2 AQA tommorow and sequences/series is my weekpoint. Am pretty good at everything else!


Your given the forumulas for sequences and series, so what i do is put them in the equations that you are given and do something to get the answer.
 
We get given a nice formula (or cheat as the teachers call it!) book with them all in :D MEI Examination Formulae and Tables (MF2) it's called (OCR MEI).

Having that book in the exam makes my two U's seem even more spectacular :(
 
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