Swapping identity

Let's generalize this. Let $m = 1001$ , $n=29$ , and $a$ , $d$ and $S_k$ be the first term, common difference and sum of first $k$ terms of the AP. Then, we have:

$\begin{cases} S_m = \dfrac{m[2a+(m-1)d]}{2} = n & \Rightarrow 2mna + mn(m-1)d = 2n^2 & ...(1) \\ S_n = \dfrac{n[2a+(n-1)d]}{2} = m & \Rightarrow 2mna + mn(n-1)d = 2m^2 & ...(2) \end{cases}$

$\begin{aligned} (1)-(2): \quad mn(m - n)d & = 2(n^2-m^2) \\ \Rightarrow \color{#3D99F6}{d} & = \color{#3D99F6}{-\frac{2(m+n)}{mn}} \end{aligned}$

Now we need to find:

$\begin{aligned} S_{m+n} & = \dfrac{(m+n)[2a+(m+n-1)d]}{2} \\ & = \dfrac{m[2a+(m-1)d + nd]}{2} + \dfrac{n[2a+(n-1)d+md]}{2} \\ & = S_m + \dfrac{mnd}{2} + S_n + \dfrac{nmd}{2} \\ & = n + m + mn\color{#3D99F6}{d} \\ & = n + m - mn\left(\color{#3D99F6}{\frac{2(m+n)}{mn}}\right) \\ & = -(m+n) \\ \Rightarrow S_{1030} & = S_{1001+29} = \boxed{- 1030} \end{aligned}$