Animated Solution for Mathematics - Sequence and Series: Suppose $a,b,c$ are in A.P. and $a^2,b^2,c^2$ are in G.P. if $a<b<c$ and $a+b+c=3/2$, then the value of $a$ is

Welcome, future engineer. Today, we are not just solving a problem about sequences; we are embarking on a journey into the heart of mathematical symmetry. When you look at a problem involving an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.) simultaneously, your first instinct might be to write down the standard definitions and start grinding through the algebra.

But wait. Stop. Take a breath. In the world of JEE Advanced, the most elegant solutions are often found by observing the structure of the problem before you touch your pen to paper.

Let us begin by visualizing our three numbers, $a,b,$ and $c$. We are told they form an A.P. and that $a<b<c$. The sum is given by:

Instead of using the standard $a,a+d,a+2d$, let us center our perspective on the middle term, $b$. By defining our terms as $b-d,b,$ and $b+d$, we invoke the power of symmetry.

When we sum these terms, the common difference $d$ vanishes into thin air:

Given the sum is $\frac{3}{2}$, we immediately find that $3b=\frac{3}{2}$, which means $b=\frac{1}{2}$. Just like that, the anchor of our progression is set. We have found the middle term without breaking a sweat.

Now, let us pivot to the second condition. We are told that $a^2,b^2,c^2$ are in G.P. This means the square of the middle term must equal the product of the outer terms:

This simplifies to $b^4=(ac{)}^2$. Here is the critical moment where many students stumble. When you take the square root of both sides, you must be rigorous. You are not just getting $b^2=ac$; you are getting $b^2=\pm ac$.

Case one is $b^2=ac$. If we substitute our expressions $a=\frac{1}{2}-d$ and $c=\frac{1}{2}+d$, we get:

This simplifies to $\frac{1}{4}=\frac{1}{4}-d^2$, which forces $d^2=0$, or $d=0$. But look back at the problem statement: $a<b<c$. If $d=0$, then $a=b=c$, which violates the strict inequality. We must reject this case.

Now, we turn to Case two: $b^2=-ac$. Substituting our values again, we get:

This becomes $\frac{1}{4}=-(\frac{1}{4}-d^2)$, which simplifies to $\frac{1}{4}=d^2-\frac{1}{4}$. Adding $\frac{1}{4}$ to both sides, we find $d^2=\frac{1}{2}$.

Since $d$ must be positive to satisfy $a<b<c$, we take the positive root: $d=\frac{1}{\sqrt{2}}$. We have arrived at the finish line.

We know $b=\frac{1}{2}$ and $d=\frac{1}{\sqrt{2}}$. The first term $a$ is simply $b-d$, which gives us:

This is the value we sought. It is a beautiful, precise result. Remember, the path to the solution was not found through brute force, but through the strategic choice of variables and a deep respect for the constraints.