
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
If \( x_n \) is an arithmetic progression, then \( x_n = b + (c - b)(n - 1) \). Then:
\[ x_{n+1} = b + (c - b) n \]
\[ x_{n+2} = b + (c - b)(n + 1) \]

Substitute these expressions into the given condition:
\[ b + (c - b)(n + 1) = 3(b + (c - b) n) - 2(b + (c - b)(n - 1)) + a \]

This simplifies to:
\[ (c - b)(n + 1 - 3n + 2(n - 1)) + a = 0 \]
Hence,
\[ a = (c - b) < 0 \]

If \( x_n \) is a geometric progression, then \( x_n = b \cdot q^{n-1} \), \( x_{n+1} = b \cdot q^n \), \( x_{n+2} = b \cdot q^{n+1} \). Substitute these expressions into the given condition:
\[ bq^{n+1} = 3 b q^n - 2 b q^{n-1} + a \]
\[ b \cdot q^{n-1}(q^2 - 3q + 2) = a \]

\( a \) being independent of \( n \) is possible when \( q = 1 \) or \( q = 2 \), resulting in \( a = 0 \). With \( q = 1 \), the progression cannot be increasing. With \( q = 2 \), the progression \( x_n \) is increasing if \( c = 2b > 0 \).

Present the answer in LaTex format: \boxed{Your answer}
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