More C1 Series problems!

Original post by nikki13

So, although I haven't missed a single lesson or anything of maths, I've been taught hardly anything about series. Most of the simple questions I could do from the textbook, but the harder problems? No chance :'( Nothing in my textbook to help or anything.

Here is what I'm reallyyyy stuck on:

Three consecutive terms of an arithmetic series are (7k-1) (5k+3) (4k+1) respectively.
a) Find the constant k
b)find the smallest possible term in the series
Given also that the series has r positive terms:
c)show that the sum of the positive terms in the series is given by r(4r-3)

I don't even know where to begin at all, I haven't started, just stared at it for a while and tried not to cry...please help!

Thanks so much!

Hopefully you can work out the value of k with the hints given.

Once you know k (and hence the 3 terms you are given) you can work out the common difference d in the terms.

for (b), did you mean "smallest possible" or "smallest positive"?

For (c), if you know the "smallest positive" term in the series, you can just sum r terms of the series, starting with this term, using the standard formula for sum of an AP.

Reply 4

OP

Ok, thanks for everything so far! I've got k=6, and worked out my terms, so they are 41,33,25 with a common difference of -8. Is that right so far?

I did mean smallest positive! Stupid me...
And yeah, now I'm stuck on b and c. I'm hopeless at this :frown:

Reply 5

Original post by nikki13

Ok, thanks for everything so far! I've got k=6, and worked out my terms, so they are 41,33,25 with a common difference of -8. Is that right so far?

I did mean smallest positive! Stupid me...
And yeah, now I'm stuck on b and c. I'm hopeless at this :frown:

Surely you can see that the numbers are decreasing so they are going to be negative after a while

Carry the series on and you will easily find the smallest positive

Reply 6

OP

Well yes, but I've got to find a formula to do it with - I can see that it's the 6th term by counting it out, but the purpose is to find a way to do it that would always work.
I'm thinking it's an inequality, I did:
41+(n-1)-8>0
Which came out to n>6.125
But I'm not sure that's right, because I know it's the 6th term, so have I got the inequality signs wrong?

Reply 7

Original post by nikki13

Well yes, but I've got to find a formula to do it with - I can see that it's the 6th term by counting it out, but the purpose is to find a way to do it that would always work.

No it isn't, answering the question is enough

I'm thinking it's an inequality, I did:
41+(n-1)-8>0

I am not really sure what you meant by that

Did you mean

41+(n-1)(-8)>0

If so I do not see how you got your answer

Reply 8

OP

Sorry, I suppose that made no sense...I used the general formula for a series, Un=a+(n-1)d and substituted in what I had - the first term, 41, and the common difference I found, -8. Then, because it has to be positive, so more than zero, I put in the inequality. So, 41+(n-1)(-8)>0

I'm always being told by my tutors that I need to find the method and how to apply it to everything - just answering the question is not enough. I don't know if they're just trying to get a bit more out of us, though.

Reply 9

Original post by nikki13

Sorry, I suppose that made no sense...I used the general formula for a series, Un=a+(n-1)d and substituted in what I had - the first term, 41, and the common difference I found, -8. Then, because it has to be positive, so more than zero, I put in the inequality. So, 41+(n-1)(-8)>0

I'm always being told by my tutors that I need to find the method and how to apply it to everything - just answering the question is not enough. I don't know if they're just trying to get a bit more out of us, though.

Ok

so your inequality is now correct but your solution to it is still incorrect

Reply 10

OP

Oh, I get it! I divided by negative eight, so the inequality changes direction! Totally forgot that rule...

Thanks for your patience with my inability to cope with simple things...clearly need a break from this :P

Reply 11

Original post by nikki13

Oh, I get it! I divided by negative eight, so the inequality changes direction! Totally forgot that rule...

Thanks for your patience with my inability to cope with simple things...clearly need a break from this :P

Reply 12

Original post by nikki13

Oh, I get it! I divided by negative eight, so the inequality changes direction! Totally forgot that rule...

Thanks for your patience with my inability to cope with simple things...clearly need a break from this :P

Actually you probably don't need to go overboard with a formula for this - if you note that your largest term is 41 and that you're constantly subtracting 8 then you just need to find the remainder when 41 is divided by 8 - this will tell you what your smallest positive term is!

Have you attempted the last part yet?

Reply 13

OP

I did...no clue, don't even know how to really start it. So now I'm kinda worried I got the rest of it wrong...

I said that since the sum is given by (n/2)(2a+(n-1)d)
You substitute it what you know - but not the 6, because you need to find it in terms of r, so I did:
(r/2)(2a+(r-1)d)
=(r/2)(2x41-8r-8)
=(r/2)(74-8r)

[Common]
The 74 seemed too high, I think I need to factorise somehow but the fraction at the front has me stuck.

Reply 14

Original post by nikki13

I did...no clue, don't even know how to really start it. So now I'm kinda worried I got the rest of it wrong...

I said that since the sum is given by (n/2)(2a+(n-1)d)
You substitute it what you know - but not the 6, because you need to find it in terms of r, so I did:
(r/2)(2a+(r-1)d)
=(r/2)(2x41-8r-8)
=(r/2)(74-8r)

[Common]
The 74 seemed too high, I think I need to factorise somehow but the fraction at the front has me stuck.

I think the trick with this part is to sum the series "in reverse order". That is, you should have found the smallest positive term to be 1. You are told that you are summing "r positive terms". You know what the common difference is (remember we going upwards from 1, so treat the difference as positive now).

Now plug these values into the general formula for the sum of an AP and see what you get :smile:

Reply 15

OP

Awesome, all sorted! Thank you so much...it's all so simple when someone else shows you how! I've checked all my notes, we haven't done anything about this kind of stuff at all, and my textbook is next to useless! Would have been so stuck without you guys!

Thank you all :biggrin:

Reply 16

Original post by nikki13

Awesome, all sorted! Thank you so much...it's all so simple when someone else shows you how! I've checked all my notes, we haven't done anything about this kind of stuff at all, and my textbook is next to useless! Would have been so stuck without you guys!

Thank you all :biggrin:

No problem - glad you got it sorted! I don't think the AP questions come more difficult than this - I had to think a bit myself before working out the right strategy for the last part :smile:

Reply 17

OP