My 600 followers problem!

Let $2015=q \quad 600=p\quad a_1=a \quad b_1=b$

$a_p=a+(p-1)d_1=q …(1)$

$a_q=a+(q-1)d_1=p …(2)$

$(2)-(1) \Rightarrow q-p = d_1(p-1-q+1) \Rightarrow q-p = d_1(p-q)\Rightarrow d_1=-1$

Substitute $d_1=-1$ in $1$ to get $q=a-p+1 \Rightarrow p+q=a+1…(3)$

$a_n=a+(n-1)d_1=a-n+1\Rightarrow a_n=p+q-n…(4)$

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$S_p=\dfrac{p[2b+(p-1)d_2]}{2}=\dfrac{2bp+p^2d_2-pd_2}{2}=q\\\Rightarrow S_q=\dfrac{q[2b+(q-1)d_2]}{2}=\dfrac{2bq+q^2d_2-qd_2}{2}=p$

$q-p =\dfrac{2bp+p^2d_2-pd_2}{2}-\dfrac{2bq+q^2d_2-qd_2}{2}\\\Rightarrow-2(p-q)=2b(p-q)+d_2(p+q)(p-q)-d(p-q) \\\Rightarrow2b+d_2(p+q)-d_2=-2\\\Rightarrow2b+d_2(p+q-1)=-2\\\Rightarrow\dfrac{(p+q)[2b+d_2(p+q-1)]}{2}=-2\left(\dfrac{(p+q)}{2}\right)$ $\Rightarrow S_{p+q}=-(p+q)…(5)$

Adding $(4),(5)$ , we get $\large a_n+S_{p+q}=-n$

$\huge\boxed{a_{1729}+S_{2615}=\color{#3D99F6}{-1729}}$

Since AP is linear,

$\dfrac {a_{1729}-a_{600}}{1729-600} = \dfrac {a_{2015}-a_{600}}{2015-600} \quad \Rightarrow \dfrac {a_{1729}-2015}{1729-600} = \dfrac {600-2015}{2015-600} \\ \Rightarrow \dfrac {a_{1729}-2015}{1129} = -1 \quad \Rightarrow a_{1729} = -1129 + 2015 = 886$

We know that:

$S_n = \dfrac {n}{2} [2a+(n-1)d] = \dfrac {d}{2} n^2+ \left(a-\dfrac{d}{2} \right)n$

Therefore,

$\begin{cases} S_{2015} & = 2015^2\dot{}\dfrac {d}{2} & - 2015\left(a - \dfrac {d}{2} \right) & = 600 &...(1) \\ S_{600} & = 600^2\dot{}\dfrac {d}{2} & - 600\left(a - \dfrac {d}{2} \right) & = 2015 &...(2) \\ S_{2615} & = 2615^2\dot{}\dfrac {d}{2} & - 2615\left(a - \dfrac {d}{2} \right) & &...(3) \end{cases}$

$\begin{cases} Eq.3-Eq.1-Eq.2: & S_{2615} = 1209000d + 2615 &...(4) \\ 600\dot{}Eq.1-2015\dot{}Eq.2: & 1710735000d = -3700225 & \\ & \Rightarrow 1209000d = -5230 &...(5) \end{cases}$

Substituting $Eq.5$ into $Eq.4: \space S_{2615} = -5236 + 2615 = \boxed{-2615}$ .