Animated Solution for Mathematics - Sequence and Series: The $4^{th}$ 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

Imagine you are standing on the edge of a vast mathematical landscape. You are looking at a sequence of numbers, a Geometric Progression (GP), where each term is born from the previous one by a simple, constant multiplication. It is a structure of perfect self-similarity.

Today, we are going to decode a specific GP, not by brute force, but by uncovering the hidden relationships between its terms and sums.

Decoding the Foundation

We begin with the basics. We have a first term, which we call $a$, and a common ratio, $r=\frac{1}{m}$, where $m$ is a natural number. The problem gives us a lighthouse in this fog: the fourth term, $T_4$, is exactly $500$.

Using the general formula for the $n^{\text{th}}$ term, $T_n=a{r}^{n-1}$, we can write:

Substituting our ratio, we get $a(\frac{1}{m}{)}^3=500$. With a quick algebraic maneuver, we isolate $a$:

This is our master key. Every term in this sequence is now tethered to the variable $m$.

The Sum Trap

Now, the problem introduces two conditions involving the sums $S_n$: $S_6>S_5+1$ and $S_7<S_6+\frac{1}{2}$. Many students will immediately reach for the sum formula $S_n=\frac{a(1-r^n)}{1-r}$.

Stop! Breathe. In JEE Advanced, the most elegant path is often the one that avoids unnecessary complexity. Remember the definition of a sum: $S_n$ is the sum of the first $n$ terms.

Therefore, the difference between $S_n$ and $S_{n-1}$ is simply the $n^{\text{th}}$ term itself:

This realization transforms our inequalities into something much more manageable:

The Inequality Dance

Now, we substitute our expressions for $T_n$ using $a=500m^3$ and $r=\frac{1}{m}$. For the sixth term:

Our first condition, $T_6>1$, becomes $\frac{500}{m^2}>1$, which simplifies beautifully to $m^2<500$. Since $m$ is a natural number, $m<\sqrt{500}\approx 22.36$.

Next, we tackle the seventh term:

Our second condition, $T_7<\frac{1}{2}$, becomes $\frac{500}{m^3}<\frac{1}{2}$. Cross-multiplying gives $m^3>1000$, which means $m>10$.

We have arrived at our destination. The valid range for $m$ is $10<m<22.36$.

Since $m$ must be a natural number, we are looking for integers in the set $\{11,12,13,\dots ,22\}$. To find the total number of values, we calculate $22-11+1$, which equals 12.

You have successfully navigated the constraints and found the solution. This is the power of logical reduction—turning a complex problem into a simple, elegant count.