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There is a sequence of numbers such that every entry except for the first entry is the arithmetic mean of its two neighboring entries. The 27th entry is 94 and the 94th entry is 39. What's the first entry?

 Jun 10, 2021
 #1
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Let $s_0, s_1, s_2, \ldots$ be the sequence we are seeking.  Write $s_0=a$ and $s_1=b$.  By the arithmetic mean assumption, we have that $2s_{n-1} = s_n + s_{n-2}$  if $n \ge 2$.  Thus, we may define the sequence as
\[
s_n =
\left\{
  \begin{array}{ll}
    a & \mbox{if $n = 0$}  \\
    b & \mbox{if $n = 1$}  \\
    2s_{n-1} - s_{n-2}  & \mbox{if $n \ge 2$.}
  \end{array}
\right.
\]

 

We claim that $s_n$ has the closed-form formula $s_n = nb - (n-1)a$.  We prove this by induction.  When $n=0$ the conjectured formula gives $s_0 = 0\cdot b - (-1)a = a$ which is correct.  When $n=1$ the conjectured formula gives $s_1 = 1\cdot b - (0)a = b$ which is also correct.  Now let $n\ge2$ and assume by induction that the conjecture formula is correct for all $k$ such that $0\le k < n$.  We have by its recursive definition and the inductive hyphothesis that 
\begin{eqnarray*}
s_n &=& 2s_{n-1} - s_{n-2} \\
    &=& 2((n-1)b-(n-2)a) - ((n-2)b - (n-3)a) \\
    &=& (2(n-1)-(n-2))b - (2(n-2)-(n-3))a \\
    &=& nb - (n-1)a
\end{eqnarray*}
and it agrees with the formula again.  Thus the conjectured formula is correct by induction.

 

By assumption $s_{27} = 94$ and $s_{94}= 39$.  This gives 2 equations $27b-26a=94$ and $94b-93a=39$.  Solving for $a$ we get $a=\frac{7783}{67}$.  By the way, $b=\frac{7728}{67}$ but the question didn't ask for it.
 

 Jun 12, 2021

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