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

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1. "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.

Thanks!
2. Let 99 be the nth term.

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

So 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)?
4. (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. is the nth term, whereas 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,

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