Summation of a Sequence using previous term You have a data table going from 1st term to 1000th term.
The 1st term is -6 and in the second third of the data table you see a term that gives 1.
From 2nd term to 999th term the value equals the sum of the value of the term right before and right after it.
Which of the following statements is true?
1 - The value of the 991st term is of 1. 
2 - The value of the 992nd term is of 2. 
3 - The value of the 993rd term is of 3. 
4 - The value of the 994th term is of 4. 
5 - The value of the 995th term is of 5. 
6 - The value of the 996th term is of 6. 
7 - The value of the 997th term is of 7. 
8 - The value of the 998th term is of 8. 
9 - The value of the 999th term is of 9. 
10 - The value of the 1000th term is of 10. 
So I thought this would be fairly straightforward thinking nth term = sum of n-1 term and n+1 term. Issue is Each term can be made to have infinite 1st terms and that gives an endless summation of -6 which is going negative and not positive. 
Does anyone have an alternative suggestion on how to approach this?

Starting off with our formula $a_n = a_{n-1} + a_{n+1}$, I expanded out the $a_{n-1}$ to get
$$a_n = a_{n-2} + a_n + a_{n+1} \implies a_{n+1} = -a_{n-2}$$
so now our sequence looks like $-6, a_2, a_3, 6, a_5, a_6, -6, \dots$. Since $991 \equiv 1 \mod 6$, $a_{991} = -6$, $a_{994} = 6$, and $a_{997} = -6$. None of those match your statements, so you'll need to use more information given to continue.
Spoilers for a bigger hint:

 Let $a_3 = n$. This determines the rest of the sequence as $$\begin{array}. a_{6k} &= -n\\ a_{6k+1} &= -6\\ a_{6k+2} &= n-6\\ a_{6k+3} &= n\\ a_{6k+4} &= 6\\ a_{6k+5} &= 6-n\\ \end{array}$$ Now we have 4 possibilities: either $-n=1$, $n-6=1$, $n=1$, or $6-n=1$, so we just need to check which of these satisfies one of those ten statements.

