Interesting sequence

Since $a_1 = a_{201} = 19999$ and $a_2=\dfrac {19999+a_3}2 - d$ and $a_{200}=\dfrac {19999+a_{199}}2 - d$ , we deduce that $a_2=a_{200}$ , $a_3=a_{199}$ , $a_4 = a_{198}$ , $\cdots$ $\implies a_k = a_{202-k}$ $\implies a_{100} = a_{102}$ and $a_{101} = \dfrac {a_{100}+a_{102}}2-d = a_{100}-d$ . From:

$\begin{aligned} a_n & = \frac {a_{n-1}+a_{n+1}}2 - d \\ \implies a_{n-1} & = 2a_n - a_{n+1} + 2d & \small \color{#3D99F6} \text{Putting }n=100 \\ a_{99} & = 2a_{100} - {\color{#3D99F6}a_{101}} + 2d & \small \color{#3D99F6} \text{Since }a_{101} = a_{100} - d \\ & = a_{100}+3d \\ a_{100} & = a_{100} \\ a_{101} & = a_{100} - d \\ \implies a_{102-k} & = a_{100} + k(k-2) d & \small \color{#3D99F6} \text{See proof} \\ a_1 & = a_{100} + 101(101-2)d & \small \color{#3D99F6} \text{Putting }k=101 \\ 19999 & = a_{100} + 9999d \\ \implies d & = 1 & \small \color{#3D99F6} \text{Only possible positive integer solution} \\ a_{100} & = 10000 \\ \implies a_{102-k} & = 10000 + k(k-2) & \small \color{#3D99F6} \text{Putting }k=76 \\ a_{26} & = 10000 + 76(76-2) \\ & = \boxed{15624} \end{aligned}$

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Proof: Proof by induction the claim $a_{102-k} = a_{100} + k(k-2)d$ for all $k \ge 1$ .

For $k=1$ , $a_{101} = a_{100} + 1(1-2) = a_{100} - d$ and for $k=2$ , $a_{100} = a_{100} + 2(2-2) = a_{100}$ , implying the claim is true for $k=1$ and $k=2$ . Assuming that the claim is true for $k$ and $k+1$ , then:

$\begin{aligned} a_{102-(k+2)} & = 2a_{102-(k+1)} - a_{102-k} + 2d \\ & = 2(a_{100} + (k+1)(k-1)d) - (a_{100} + k(k-2)d) + 2d \\ & = a_{100} + (k^2+2k)d \\ & = a_{100} + (k+2)(k+2-2)d \end{aligned}$

Implying that the claim is also true for $k+2$ and hence true for all $k > 1$ .

First show that $a_n = c + b n + a n^2$ for some constants $a,b,c$ , using the standard method for solving linear recurrence relations. Then, solve for $a,b,c$ in terms of $d,a_2$ (using $a_1 = 19999$ ). Then use the condition $a_{201} = 19999$ to solve for $a_2$ in terms of $d$ . Then you have a formula for $a_n$ dependent on $d$ . It is easy to check that $a_n>0$ only if $d=1$ .