Q regarding finding sum of first 2002 terms Q : A sequence of integers $a_{1}+a_{2}+\cdots+a_{n}$ satisfies $a_{n+2}=a_{n+1}$ $-a_{n}$ for $n \geq 1$. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. Find the sum of the first 2002 terms.
My questions regarding this problem are:

*

*What will be the 1st term ?

Since they say n ≥1. So , do we say 1st term of the sequence has n = 2.
$a_{2+2} = a_{2+1} - a_2$


*It will be great if u could please share different-different ways u can solve it. Also , a very important point is that : What are you thinking at every step while solving this Q. This really helps me a lot because then I can also know what is the way you’re trying to find the solution. Like if it’s a method you’re using , did you already know about it ? Or you’re thinking of ways to find the solution i.e how are u thinking to find ways to solve ?

3)What I have tried:

*

*Sum of 1000th - 1003th term:

1003 - ( -999) = 2002.

*

*What we need to find is the sum of next 999terms. I’m not able to solve further than that.



*Also , please share if you what are the other method that we can use ?

Thank you.

M:1 = Telescoping
I can see we can’t do anything after this. So , I use telescoping method w/ the help of user dxiv.
$\begin{aligned} a_{n+2} &=a_{n+1}-a_{n} \\ \therefore \quad a_{n+3} &=a_{n+2}-a_{n+1} \\ &=a_{n+1}-a_{n}-a_{n+1} \\ &=-a_{n} \end{aligned}$ $$ \begin{array}{ll}a_{n+4} & =a_{n+3}-a_{n+2} \\ \therefore \quad a_{n+5} & =a_{n+4}-a_{n+3} \\ \therefore \quad a_{n+4} & +a_{n+5}=a_{n+4}-a_{n+2} \\ \therefore \quad a_{n+5} & =-a_{n+2} \\ \therefore \quad a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5} & \\ & =a_{n}+a_{n+1}+a_{n+2}-a_{n}+a_{n+3}-a_{n+2}-a_{n+2} \\ & = & a_{n+1}+a_{n+3}-a_{n+2}=0\end{array} $$
Let $S_{n}$ denotes the sum of first $n$ terms
$$
\begin{aligned}
S_{999} &=S_{6 \times 166+3}=S_{3} &(\because \text { every '6' consecutive }\\
S_{1003} &=S_{6 \times 167+1}=S_{1}^{\circ} &\text { terms has sum zero. }) \\
S_{2002} &=S_{6 \times 333+4}=S_{4} & \\
S_{2002} &=S_{4}=a_{1}+a_{2}+a_{3}+a_{4} \\
&=S_{3}-a_{1} \\
&=1003-(-999) \\
&=2002
\end{aligned} \quad\left(\because a_{4}=-a_{1} \text { since } a_{n+3}=-a_{n}\right)
$$
