Key Concepts and Formulas

• Geometric Progression (GP): A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (r).
  • $n$-th term: $T_n = ar^{n-1}$, where 'a' is the first term.
  • Sum of the first $n$ terms: $S_n = \frac{a(1 - r^n)}{1 - r}$, for $r \neq 1$.
• Relationship between Sums and Terms: The difference between consecutive sums of a GP is equal to the term at the higher index: $S_n - S_{n-1} = T_n$.

Step-by-Step Solution

Step 1: Define the GP and use the given information to find the first term 'a'.

We are given a Geometric Progression (GP) with the 4th term $T_4 = 500$ and a common ratio $r = \frac{1}{m}$, where $m \in \mathbb{N}$ (natural numbers). The formula for the $n$-th term of a GP is $T_n = ar^{n-1}$. For the 4th term, we have $T_4 = ar^{4-1} = ar^3$. Substituting the given values: $ar^3 = 500$ Now, substitute $r = \frac{1}{m}$: $a\left(\frac{1}{m}\right)^3 = 500$ $a \cdot \frac{1}{m^3} = 500$ Solving for 'a', we get: $a = 500m^3$

Step 2: Analyze the first inequality: $S_6 > S_5 + 1$.

The inequality $S_6 > S_5 + 1$ can be rewritten by moving $S_5$ to the left side: $S_6 - S_5 > 1$ We know that $S_n - S_{n-1} = T_n$. Therefore, $S_6 - S_5$ is the 6th term of the GP, $T_6$. So, the inequality becomes: $T_6 > 1$ The formula for the 6th term is $T_6 = ar^{6-1} = ar^5$. Substitute the expressions for 'a' and 'r': $a = 500m^3$ $r = \frac{1}{m}$ $T_6 = (500m^3)\left(\frac{1}{m}\right)^5$ $T_6 = 500m^3 \cdot \frac{1}{m^5}$ $T_6 = \frac{500}{m^2}$ Now, apply the inequality $T_6 > 1$: $\frac{500}{m^2} > 1$ Since $m \in \mathbb{N}$, $m^2$ is positive, so we can multiply both sides by $m^2$ without changing the inequality direction: $500 > m^2$ $m^2 < 500$ To find the possible integer values of 'm', we take the square root of both sides: $m < \sqrt{500}$ We know that $22^2 = 484$ and $23^2 = 529$. So, $\sqrt{500}$ is between 22 and 23. $m < 22.36...$ Since $m$ must be a natural number ($m \in \mathbb{N}$), the possible values for $m$ from this inequality are $1, 2, 3, ..., 22$.

Step 3: Analyze the second inequality: $S_7 < S_6 + \frac{1}{2}$.

The inequality $S_7 < S_6 + \frac{1}{2}$ can be rewritten by moving $S_6$ to the left side: $S_7 - S_6 < \frac{1}{2}$ Similarly, $S_7 - S_6$ is the 7th term of the GP, $T_7$. So, the inequality becomes: $T_7 < \frac{1}{2}$ The formula for the 7th term is $T_7 = ar^{7-1} = ar^6$. Substitute the expressions for 'a' and 'r': $a = 500m^3$ $r = \frac{1}{m}$ $T_7 = (500m^3)\left(\frac{1}{m}\right)^6$ $T_7 = 500m^3 \cdot \frac{1}{m^6}$ $T_7 = \frac{500}{m^3}$ Now, apply the inequality $T_7 < \frac{1}{2}$: $\frac{500}{m^3} < \frac{1}{2}$ Since $m \in \mathbb{N}$, $m^3$ is positive, so we can multiply both sides by $2m^3$ without changing the inequality direction: $500 \cdot 2 < m^3$ $1000 < m^3$ $m^3 > 1000$ To find the possible integer values of 'm', we take the cube root of both sides: $m > \sqrt[3]{1000}$ $m > 10$ Since $m$ must be a natural number, the possible values for $m$ from this inequality are $11, 12, 13, ...$.

Step 4: Combine the conditions on 'm' to find the number of possible values.

From Step 2, we found that $m$ must be a natural number such that $m \leq 22$. So, $m \in \{1, 2, ..., 22\}$. From Step 3, we found that $m$ must be a natural number such that $m > 10$. So, $m \in \{11, 12, 13, ...\}$.

To satisfy both conditions, $m$ must be a natural number that is greater than 10 AND less than or equal to 22. The set of possible values for $m$ is $\{11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22\}$.

Step 5: Count the number of possible values of 'm'.

The number of integers in the set $\{11, 12, ..., 22\}$ is calculated as (Last Value - First Value + 1). Number of values = $22 - 11 + 1 = 11 + 1 = 12$.

Common Mistakes & Tips

• Algebraic Errors: Be careful with exponent rules and fraction manipulation, especially when substituting $r = \frac{1}{m}$.
• Inequality Direction: Ensure the direction of the inequality remains correct after multiplication or division by variables. Since $m \in \mathbb{N}$, $m$ and its powers are always positive.
• Integer Constraints: Always remember that $m$ must be a natural number. This restricts the possible values of $m$ to positive integers.
• Range Calculation: When counting the number of integers in an inclusive range $[a, b]$, the formula is $b - a + 1$.

Summary

We used the properties of geometric progressions to express the first term 'a' in terms of 'm'. Then, we simplified the given inequalities by recognizing that the difference between consecutive sums equals the corresponding term. This led to two inequalities involving 'm': $m^2 < 500$ and $m^3 > 1000$. By solving these inequalities and considering that 'm' must be a natural number, we determined the range of possible values for 'm' and counted them to find the final answer.

The final answer is $\boxed{12}$.