Animated Solution for Mathematics - Sequence and Series: Suppose $a,b,c$ are in A.P. and $a^2,2b^2,c^2$ are in G.P. If $a<b<c$ and $a+b+c=1$, then $9(a^2+b^2+c^2)$ is equal to ......... .

Imagine you are standing on the edge of a mathematical landscape where numbers don't just exist; they dance in perfect, predictable patterns. Today, we are exploring the elegant interplay between Arithmetic Progressions (A.P.) and Geometric Progressions (G.P.).

We start with three numbers, $a,b,$ and $c$, locked in an Arithmetic Progression. The moment you see this, your mind should immediately jump to the definition of an A.P.: the difference between consecutive terms is constant.

This gives us the beautiful symmetry $b-a=c-b$, which simplifies to:

We are also given the sum $a+b+c=1$. By substituting $a+c=2b$ into this sum, we get $2b+b=1$, which leads us directly to $3b=1$, or $b=\frac{1}{3}$.

Just like that, we have anchored our sequence. We know the middle term, and we know that $a+c=\frac{2}{3}$. This is our first major victory.

Trap

Now, the problem introduces a twist: $a^2,2b^2,c^2$ are in a Geometric Progression. The fundamental property of a G.P. is that the square of the middle term equals the product of the extremes.

So, we write:

Expanding this, we get $4b^4=(ac{)}^2$. Since we already know $b=\frac{1}{3}$, we substitute it:

Thus, $(ac{)}^2=\frac{4}{81}$. Taking the square root, we find two potential paths: $ac=\frac{2}{9}$ or $ac=-\frac{2}{9}$. This is where many students stumble.

To decide between our two candidates for $ac$, we must look at the constraint $a<b<c$. This implies that $a$ and $c$ must be real numbers.

Let's form a quadratic equation $x^2-(a+c)x+ac=0$ whose roots are $a$ and $c$. If we test $ac=\frac{2}{9}$, the equation becomes:

The discriminant $D=B^2-4AC$ here is:

A negative discriminant means the roots are imaginary! But our constraint demands real numbers. Therefore, we must reject $ac=\frac{2}{9}$. The only valid path is $ac=-\frac{2}{9}$.

We have arrived at the final stage. We need to find $9(a^2+b^2+c^2)$. We use the algebraic identity $a^2+c^2=(a+c{)}^2-2ac$.

Adding $b^2$ to both sides, we get $a^2+b^2+c^2=(a+c{)}^2-2ac+b^2$. Substituting our known values:

This becomes:

Finally, multiplying by $9$, we get $9(1)=9$. The elegance of this result is a testament to the beauty of mathematics. The final answer is 9.