In a sequence of positive integers each term after the first is $\frac{1}{3}$ of the sum of the term that precedes it and the term that follows it in the sequence. What is the 5th term of this sequence if the 1st term is 2 and the 4th term is 34?
Let $a,b,c$ be 2nd, 3rd, and 5th terms respectively. Our sequence is therefore $2,a,b,34,c,\dots$. From the information given, we have \begin{align*}
a &= \frac13(2+b)\\
b &= \frac13(a+34)\\
34 &= \frac13(b+c).
\end{align*} Before we find $c$, we use the first two equations to solve for $b$. Substituting $a = \frac13(2+b)$, we obtain \begin{align*}
b &= \frac13(\frac13(2+b)+34)\\
\Rightarrow 3b &= \frac13(2+b)+34\\
\Rightarrow 9b &= 2+b+102\\
\Rightarrow 8b &= 104\\
\Rightarrow b &= 13.
\end{align*} Substituting $b = 13$ into $34 = \frac13(b+c)$, we obtain \begin{align*}
34 &= \frac13(13+c)\\
\Rightarrow 102 &= 13+c\\
\Rightarrow c &= \boxed{89}.
\end{align*}
89
Level 5
Algebra
Problem: In a sequence of positive integers each term after the first is $\frac{1}{3}$ of the sum of the term that precedes it and the term that follows it in the sequence. What is the 5th term of this sequence if the 1st term is 2 and the 4th term is 34?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the 5th term in the sequence, we denote the terms as follows: let \( a_1 = 2 \), \( a_2 \), \( a_3 \), \( a_4 = 34 \), and \( a_5 \). The relationship given in the problem states that each term after the first is one-third of the sum of the term that precedes it and the term that follows it. This can be expressed mathematically as:

\[
a_n = \frac{1}{3}(a_{n-1} + a_{n+1})
\]

Rearranging this equation gives:

\[
3a_n = a_{n-1} + a_{n+1}
\]

From this, we can derive the following relationships for the terms in the sequence:

1. For \( n = 2 \):
   \[
   3a_2 = a_1 + a_3 \implies 3a_2 = 2 + a_3 \implies a_3 = 3a_2 - 2
   \]

2. For \( n = 3 \):
   \[
   3a_3 = a_2 + a_4 \implies 3a_3 = a_2 + 34 \implies a_2 = 3a_3 - 34
   \]

3. For \( n = 4 \):
   \[
   3a_4 = a_3 + a_5 \implies 3 \cdot 34 = a_3 + a_5 \implies 102 = a_3 + a_5 \implies a_5 = 102 - a_3
   \]

Now, we have two equations involving \( a_2 \) and \( a_3 \):

1. \( a_3 = 3a_2 - 2 \)
2. \( a_2 = 3a_3 - 34 \)

Substituting the expression for \( a_3 \) into the second equation:

\[
a_2 = 3(3a_2 - 2) - 34
\]
\[
a_2 = 9a_2 - 6 - 34
\]
\[
a_2 = 9a_2 - 40
\]
\[
40 = 8a_2 \implies a_2 = 5
\]

Now substituting \( a_2 \) back to find \( a_3 \):

\[
a_3 = 3(5) - 2 = 15 - 2 = 13
\]

Next, we can find \( a_5 \) using the equation derived from \( n = 4 \):

\[
a_5 = 102 - a_3 = 102 - 13 = 89
\]

Thus, the 5th term of the sequence is \( 89 \).<|END_OF_LATENT|> The final answer is $89$.