# Find the sum of the $n$-th term of : “$-6,-12,18,24,-30,-36,42,48,\cdots$”

Consider the sequence $$-6,-12,18,24,-30,-36,42,48,\cdots$$

If the sum of first $n$ terms of the sequence is $120$,how to find $n$?

Each term is a increasing multiple of $6$ but I am not sure how to take care of $\pm$ which repeates after two terms, any ideas?

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Try splitting it into (the sum of) four sequences, two with negative terms and two with positive terms. – The Chaz 2.0 Dec 12 '11 at 16:20
In any event: check the sequence generated by $$-3 \left((2k+3)\sin\left(\frac{\pi k}{2}\right)+\cos\left(\frac{\pi k}{2}\right)+1\right)$$ – J. M. Dec 12 '11 at 16:31
Can you guess the rule for the absolute values of the numbers? And for their signs? – Marc van Leeuwen Dec 12 '11 at 16:33
@mixedmath: it might be too late, and I'm mobile (so it's tricky to mark it all up)! – The Chaz 2.0 Dec 12 '11 at 16:33

HINT $\$ Cancel a factor of $6$ then note $\rm\ 4k\ - (4k+1) -(4k+2)\ +\ 4k+3\ =\ -1 -2 + 3\ =\ 0$ hence the sum up to the term $\rm\:4\:n\:$ is simply $\rm\: 4\:n\:,\:$ since all prior terms are chunks of $4$ that sum to $0$. Thus we obtain the sum $\rm\ 4\:n = 120/6 = 20\:$ for $\rm\: n\ =\ \ldots$
$n=5$, hence there will be $20$ terms, which is the answer. – Quixotic Dec 12 '11 at 18:15
Hint: The sum of each $4$ consecutive terms (beginning with $n\equiv 0 \pmod 4$) is $24$, since $-n + -(n+6) + (n+12) + (n+18) = 24$. So the sum of the first $4n$ terms is $24n$.