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C1 - Question for very basic arithmetic sequence - how is "n" this value? Watch

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    "Find trhe sum of the multiples of 3 less than 100"

    So far, I got: a=3, d=3 but don't know how to work out "n"

    Writing out the sequence, I get 3 +6 +9 +12...+96+99

    I can't work out "n" though to put it in the equation of:

    Sn = n/2 [2a + (n-1)d]

    Looking at the solution bank, i see the value of "n" is [(99-3)/3]+1, which =33, shows that there are 33 terms in the sequence.

    How do you work out that "n" = [(99-3)/3]+1 ?

    I get that the 99 would be the last multiple of 3 less than 100, but I dont get why it is 99-3, and then (99-3)/3 (i dont get why that is divided by 3). I also don't get why there is a +1 at the end.

    Please can you explain?

    Thanks!
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    Let 99 be the nth term.

     u_n = a+(n-1)d

    So  99 = 3+ 3(n-1) and solve for n.
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    (Original post by NotNotBatman)
    Let 99 be the nth term.

     u_n = a+(n-1)d

    So  99 = 3+ 3(n-1) and solve for n.
    If you let the nth term be 99 to solve it, for Un = a+(n-1)d, wouldn't it be 99 = 3+ 3(99-1)? Because if you let the n in Un be 99, don't you have to let the n in (n-1) also be 99 when solving it?

    If you can leave the "n" in (n-1) just as "n" instead of switching it to 99 like in "Un", why so? Why is this possible? I thought if you substitute 99 for the n in Un that you would also have to substitute the (n-1) for (99-1)?
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    (Original post by blobbybill)
    If you let the nth term be 99 to solve it, for Un = a+(n-1)d, wouldn't it be 99 = 3+ 3(99-1)? Because if you let the n in Un be 99, don't you have to let the n in (n-1) also be 99 when solving it?

    If you can leave the "n" in (n-1) just as "n" instead of switching it to 99 like in "Un", why so? Why is this possible? I thought if you substitute 99 for the n in Un that you would also have to substitute the (n-1) for (99-1)?
    No, n is not the nth term it is the number of terms, which you are trying to find.  u_n is the nth term, whereas  n is the number of terms.

    If it makes it easier, you can use the kth term rather than the nth term and find the sum to k, s_k
 
 
 
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