Animated Solution for Mathematics - Sequence and Series: Does there exist a geometric progression containing $27,8$ and $12$ as three of its terms ? If it exits, how many such progressions are possible ?

To begin our journey, we must define our tools. A $G.P.$ is defined by its first term, $a$, and its common ratio, $r$. Any term in this sequence can be expressed as $T_n=a{r}^{n-1}$.

However, since we do not know the positions of our three numbers, let us be more general. Let the terms $27,8,$ and $12$ occupy positions $p,q,$ and $s$ respectively. We can write our equations as:

Here, we have used $p,q,$ and $s$ as the exponents for simplicity. This is a classic JEE strategy: when the exact position doesn't matter, use variables to represent the relative distance between terms.

Eliminating the Unknowns

We have three equations and five unknowns: $a,r,p,q,$ and $s$. At first glance, this looks like an impossible task. But look closer; the variable $a$ is a common scaling factor.

It is the 'anchor' of the sequence, but it is also the obstacle. To find the relationship between the positions, we must eliminate $a$. We do this by dividing the equations.

Let us divide the first by the second:

The $a$ cancels out, leaving us with $r^{p-q}=\frac{27}{8}$. Now, let us do the same for the third and second equations:

This simplifies to $r^{s-q}=\frac{3}{2}$. We have successfully reduced our problem to two elegant equations involving only the common ratio $r$ and the differences in positions.

The Breakthrough

Now, we look for the connection. We have $r^{p-q}=\frac{27}{8}$ and $r^{s-q}=\frac{3}{2}$. Do you see the beauty here?

$27$ is $3^3$ and $8$ is $2^3$. Therefore, $\frac{27}{8}=(\frac{3}{2}{)}^3$. This is the 'Aha!' moment. We can substitute our second equation into the first:

By the laws of exponents, this becomes $r^{p-q}=r^{3(s-q)}$. Since the bases are the same, the exponents must be equal:

A World of Possibilities

We have arrived at a linear equation: $p-q=3(s-q)$. This is a linear Diophantine equation. The variables $p,q,$ and $s$ must be integers because they represent positions in a sequence.

For any integer value we choose for $q$ and $s$, we can find a corresponding integer $p$. Because there are infinitely many integers, there are infinitely many sets of $(p,q,s)$ that satisfy this condition.

Each set corresponds to a unique common ratio $r$. Thus, not only does such a $G.P.$ exist, but there are infinitely many such progressions. You have just navigated the logic of sequences, eliminated the noise, and found the infinite structure hidden within three simple numbers.