Relation $a_{i-1}+ a_i+ a_{i+1} = 2025$ Analysis

The problem provides a sequence of integers $a_1, a_2, ..., a_{300}$ governed by the recurrence relation $a_{i-1}+ a_i+ a_{i+1} = 2025$ for $i = 2, 3, ..., 299$.

Sequence Periodicity Derivation

To understand the sequence's behavior, we analyze the recurrence relation:

• For index $i$: $a_{i-1} + a_i + a_{i+1} = 2025$
• For index $i+1$: $a_i + a_{i+1} + a_{i+2} = 2025$

Subtracting the first equation from the second gives:

$ (a_i + a_{i+1} + a_{i+2}) - (a_{i-1} + a_i + a_{i+1}) = 2025 - 2025 $

$ a_{i+2} - a_{i-1} = 0 $

$ a_{i+2} = a_{i-1} $

This equality holds for $i = 2, 3, ..., 299$. It demonstrates that the sequence is periodic with a period of 3. That is, $a_n = a_{n-3}$ for $n \ge 4$. Consequently, the value of any term $a_n$ depends only on the remainder of $n$ when divided by 3.

$a_7$ and $a_9$ Values Used

We are given specific values for two terms: $a_7 = -5$ and $a_9 = 37$. Using the periodicity ($a_n = a_{n-3}$):

• For $a_7$: Since $7 \equiv 1 \pmod 3$, $a_7 = a_{7-3} = a_4 = a_{4-3} = a_1$. Therefore, $a_1 = -5$.
• For $a_9$: Since $9 \equiv 0 \pmod 3$ (or $9 \equiv 3 \pmod 3$), $a_9 = a_{9-3} = a_6 = a_{6-3} = a_3$. Therefore, $a_3 = 37$.

$a_{106}$ Calculation

To find the value of $a_{106}$, we first determine the position of the index 106 within the cycle of 3. We calculate the remainder of 106 divided by 3:

$ 106 \div 3 = 35 \text{ with a remainder of } 1 $

This means $106 \equiv 1 \pmod 3$. Due to the sequence's periodicity, $a_{106}$ will be equal to the term $a_n$ where $n \equiv 1 \pmod 3$. Thus:

$ a_{106} = a_1 $

Final Answer

From the analysis in the previous steps, we found that $a_1 = -5$. Therefore, the value of $a_{106}$ is:

$ a_{106} = -5 $

This result matches Option C.