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Maths C1 HELP!!!!!!!!!!!!! ARITHMETIC PROGRESSIONS

An AP is given by k, 2k/3, k/3, 0, . . ..
(i) Find the sixth term.
Using a+(n-1)d I get k+5d but I don't know how to solve d.
Please help!
Original post by CounTolstoy
An AP is given by k, 2k/3, k/3, 0, . . ..
(i) Find the sixth term.
Using a+(n-1)d I get k+5d but I don't know how to solve d.
Please help!


I do not think there is a way of solving for k (without more information, though I am not doubting that you have provided the full question) but I think that they want the answer in terms of k.
Original post by CounTolstoy
An AP is given by k, 2k/3, k/3, 0, . . ..
(i) Find the sixth term.
Using a+(n-1)d I get k+5d but I don't know how to solve d.
Please help!
d is the distance between each term. In an arithmetic progression, d is constant.
Original post by SeanFM
I do not think there is a way of solving for k (without more information, though I am not doubting that you have provided the full question) but I think that they want the answer in terms of k.


You're right, I don't actually have to solve k but I don't know how to give the answer in terms of k without finding the value of d. This means I must figure out the common difference but I see no pattern!
Original post by CounTolstoy
You're right, I don't actually have to solve k but I don't know how to give the answer in terms of k without finding the value of d. This means I must figure out the common difference but I see no pattern!


There is a pattern there :tongue: but you also don't need to spot a pattern - as above, an artihmetic progression has a constant difference term 'd' (hence you use the formula for the nth term, a + (n-1)d because the first term is a, the second term you add d on (so it's a+d), the third term you add d to the second term (so it's a + d + d = a +2d).. and so on.
Original post by CounTolstoy
An AP is given by k, 2k/3, k/3, 0, . . ..
(i) Find the sixth term.
Using a+(n-1)d I get k+5d but I don't know how to solve d.
Please help!


d is the common difference between each term in the AP. So what's the difference between 2k/3 and k and k/3 and 2k/3? That's your value of d. *

For example, 5, 8, 11, 14, 17... is an AP and d=3.*
Original post by kingaaran
d is the common difference between each term in the AP. So what's the difference between 2k/3 and k and k/3 and 2k/3? That's your value of d. *

For example, 5, 8, 11, 14, 17... is an AP and d=3.*


I understand d is a constant...but I don't understand the pattern in the sequence :frown:. k ----> 2k/3 is just multiply by 2 and divide by three but that is definitely incorrect. I'm sorry but I can't understand this. Can you please explain?
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Original post by CounTolstoy
I understand d is a constant...but I don't understand the pattern in the sequence . k ----> 2k/3 is just multiply by 2 and divide by three but that is definitely incorrect. I'm sorry but I can't understand this. Can you please explain?


Common difference is −k/3 -k/3 . That is all you can do. So an=k−13(n−1)k a_n=k-\frac{1}{3}(n-1)k .
(edited 7 years ago)
Original post by SeanFM
There is a pattern there :tongue: but you also don't need to spot a pattern - as above, an artihmetic progression has a constant difference term 'd' (hence you use the formula for the nth term, a + (n-1)d because the first term is a, the second term you add d on (so it's a+d), the third term you add d to the second term (so it's a + d + d = a +2d).. and so on.


Right so, the first term will be a+(n-1)d
The second term will be a+d+(n-1)d, is that correct, Sean?
If so, may I just ask why we are multiplying the result of (n-1) by d?
My initial question remains though: how can I obtain the value of d? I don't know what sort of pattern there is, progressing from a single k to a 2k/3 to k/3.
Thank you in advance,
Original post by CounTolstoy
Right so, the first term will be a+(n-1)d
The second term will be a+d+(n-1)d, is that correct, Sean?
If so, may I just ask why we are multiplying the result of (n-1) by d?
My initial question remains though: how can I obtain the value of d? I don't know what sort of pattern there is, progressing from a single k to a 2k/3 to k/3.
Thank you in advance,


Not quite - if you reread my previous post you will see that the second term is a+2d rather than a + d + (n-1)d, and I have shown you how/why you multiply n-1 by d (where n is the term number, eg with a + (n-1)d, the first term (case where n=1), the term is a + (1-1)d = a + 0d = a.

I've also answered your second question in my previous post :h: so I am not sure what to say without repeating anything.
Original post by B_9710
Common difference is −k/3 -k/3 . That is all you can do. So an=k−13(n−1)k a_n=k-\frac{1}{3}(n-1)k .


How did you solve the common difference? Is there a method to it? I also don't understand how that IS the common difference because if we substitute your proposed value of d in the form a+(n-1)d, we get:
k+(5)(-k/3) = k-5k/15
I am quite sure I have done something wrong somewhere but I am very confused. I am not doubting your answer, I just want an explanation with regards to how you found d and why d = -k/3.
Thank you in advance!
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Original post by CounTolstoy
How did you solve the common difference? Is there a method to it? I also don't understand how that IS the common difference because if we substitute your proposed value of d in the form a+(n-1)d, we get:
k+(5)(-k/3) = k-5k/15
I am quite sure I have done something wrong somewhere but I am very confused. I am not doubting your answer, I just want an explanation with regards to how you found d and why d = -k/3.
Thank you in advance!


To find the common difference subtract the first term from the second term.
Original post by CounTolstoy
How did you solve the common difference? Is there a method to it? I also don't understand how that IS the common difference because if we substitute your proposed value of d in the form a+(n-1)d, we get:
k+(5)(-k/3) = k-5k/15
I am quite sure I have done something wrong somewhere but I am very confused. I am not doubting your answer, I just want an explanation with regards to how you found d and why d = -k/3.
Thank you in advance!


Yeah you have. Is 5*(k/3) = 5k/15?
Original post by kingaaran
Yeah you have. Is 5*(k/3) = 5k/15?


no its 5k/3

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Original post by HFancy1997


Of course it is - I know that, lol. I'm questioning their working because the OP was asking what he did wrong...
Original post by HFancy1997


Lol your genuine attempt at correcting him made me laugh
Original post by kingaaran
Of course it is - I know that, lol. I'm questioning their working because the OP was asking what he did wrong...


I just realised my mistake. Thanks very much for pointing that out.
Original post by B_9710
To find the common difference subtract the first term from the second term.


Ooooooh I never knew that. So 2k/3 -k = k/3 = d. Am I correct?
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Original post by CounTolstoy
Ooooooh I never knew that. So 2k/3 -k = k/3 = d. Am I correct?


Yes. It makes sense because a2−a1=(a+d)−a=d a_2 - a_1=(a+d)-a =d. In fact the common difference in general is an+1−an a_{n+1}-a_{n} .

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