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    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!
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    (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.
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    (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.
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    (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!
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    (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 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.
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    (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.*
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    (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 . 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 . That is all you can do. So  a_n=k-\frac{1}{3}(n-1)k .
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    (Original post by SeanFM)
    There is a pattern there 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,
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    (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 so I am not sure what to say without repeating anything.
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    (Original post by B_9710)
    Common difference is  -k/3 . That is all you can do. So  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.
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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!
    Yeah you have. Is 5*(k/3) = 5k/15?
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    (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)
    no its 5k/3

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    Of course it is - I know that, lol. I'm questioning their working because the OP was asking what he did wrong...
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    (Original post by HFancy1997)
    no its 5k/3

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    Lol your genuine attempt at correcting him made me laugh
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    (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.
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    (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  a_2 - a_1=(a+d)-a =d. In fact the common difference in general is  a_{n+1}-a_{n} .
 
 
 
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