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# Arithmetic Series Watch

1. The third and fifth terms of an arithmetic series are 48 and 45 respectively.
(a) Find the first term and the common difference of the series. (done)
(b) Find the sum of the first 8 terms of the series. (done)
The sum of the first n terms of the series is zero:
(c) Find the value of n.
$a=51$
$n=?$
$d=\frac{3}{2}$
Using: $S_{n}=\frac{n}{2}(2a+(n-1)d)$
$0=\frac{n}{2}(102+(n-1)\frac{3}{2})$
$0=\frac{n}{2}(102+\frac{3n}{2}-\frac{3}{2})$
Expand:
$0=(51n+\frac{3n^2}{4}-\frac{3n}{4})$
x4:
$0=(204n+3n^2-3n)$
Divide by 3:
$0=(68n+n^2-n)$
Simplifyyyyyyyyyyyyyy:
$0=(n^2+67n)$
Now, I could factor this and say that:
$(n+67)(n)=0$
Giving me:
$n=-67$ and $n=0$. That makes no sense. Where have I gone wrong?
2. The common difference is negative.
3. (Original post by Mr M)
The common difference is negative.
The idea of typing this out is so that I'd notice any stupid as **** mistakes that I'd make. Unbelievable. I'm an idiot. Thanks.
4. (Original post by lmsavk)
The idea of typing this out is so that I'd notice any stupid as **** mistakes that I'd make. Unbelievable. I'm an idiot. Thanks.
No problem. Sometimes it takes a second pair of eyes.

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