The 4th term of GP is 500 and its common ratio is $$\frac{1}{m}$$, $$m \in \mathbb{N}$$. Let $$S_n$$ denote the sum of the first $$n$$ terms of this GP. If $$S_6 > S_5 + 1$$ and $$S_7 < S_6 + \frac{1}{2}$$, then the number of possible values of $$m$$ is ______

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Let the first term of the Geometric Progression (GP) be $$a$$ and the common ratio be $$r = \frac{1}{m}$$, where $$m \in \mathbb{N}$$.

The $$n$$-th term of a GP is given by the formula:

$$T_n = a r^{n-1}$$

We are given that the 4th term ($$T_4$$) is $$500$$:

$$T_4 = a r^3 = 500 \implies a \left(\frac{1}{m}\right)^3 = 500 \implies a = 500m^3$$

Step 1: Analyze the first inequality

We know that the relation between the sum of first $$n$$ terms and the $$n$$-th term is:

$$S_n - S_{n-1} = T_n$$

Therefore, the first inequality $$S_6 > S_5 + 1$$ can be rewritten as:

$$S_6 - S_5 > 1 \implies T_6 > 1$$

Substituting the expressions for $$T_6$$, $a$, and $$r$$:

$$a r^5 > 1$$ $$(500m^3) \cdot \left(\frac{1}{m}\right)^5 > 1$$  $$\frac{500}{m^2} > 1 \implies m^2 < 500$$

Since $$m \in \mathbb{N}$$ and $$\sqrt{500} \approx 22.36$$, we establish the upper bound:

$$m \leq 22$$

Step 2: Analyze the second inequality

Similarly, the second inequality $$S_7 < S_6 + \frac{1}{2}$$ can be rewritten as:

$$S_7 - S_6 < \frac{1}{2} \implies T_7 < \frac{1}{2}$$

Substituting the expressions for $$T_7$$, $$a$$, and $$r$$:

$$a r^6 < \frac{1}{2}$$ $$(500m^3) \cdot \left(\frac{1}{m}\right)^6 < \frac{1}{2}$$  $$\frac{500}{m^3} < \frac{1}{2} \implies m^3 > 1000$$

Since $$10^3 = 1000$$, for $$m^3 > 1000$$ where $$m \in \mathbb{N}$$, we find the lower bound:

$$m \geq 11$$

Step 3: Combine the bounds

Combining both inequalities for $$m \in \mathbb{N}$$:

$$11 \leq m \leq 22$$

The possible values for $m$ are:

$$m \in \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22\}$$

The total number of possible values of $$m$$ is:

$$22 - 11 + 1 = 12$$

Answer:

12