Animated Solution for Mathematics - Sequence and Series: Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is:

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant interplay between two fundamental structures in mathematics: the Geometric Progression (G.P.) and the Arithmetic Progression (A.P.).

Imagine you are standing at the start of a sequence. We are given three positive numbers that form an increasing G.P. Let us define these terms as $a$, $ar$, and $a{r}^2$. Here, $a$ is our starting point, and $r$ is the common ratio that dictates the growth.

The problem provides a crucial constraint: the G.P. is increasing. For a sequence of positive numbers to grow, our common ratio $r$ must be strictly greater than $1$. Keep this inequality, $r>1$, etched in your mind as our final filter.

Now, let us introduce the twist. The problem asks us to double the middle term. Our sequence, which was $a,ar,a{r}^2$, transforms into $a,2ar,a{r}^2$.

By multiplying the middle term by $2$, we have forced it to jump. The problem states that these new numbers now form an A.P. This means the gap between the first and second term is identical to the gap between the second and third term.

How do we mathematically capture the essence of an A.P.? For any three terms $x,y,z$ in an A.P., the middle term is the arithmetic mean of the neighbors. That is, $2y=x+z$.

Let us apply this to our transformed sequence where $x=a$, $y=2ar$, and $z=a{r}^2$. Substituting these into our A.P. condition, we get:

Since we know $a>0$, we can divide both sides by $a$ without fear. This leaves us with a clean, elegant quadratic equation:

Now, we reach for our trusty quadratic formula. For an equation $A{r}^2+Br+C=0$, the roots are given by:

Plugging in our coefficients $A=1,B=-4,C=1$, we find:

Simplifying the discriminant, we get $16-4=12$. Thus, $r=\frac{4\pm 2\sqrt{3}}{2}$, which simplifies to $r=2\pm \sqrt{3}$.

We have two candidates for $r$: $2+\sqrt{3}$ and $2-\sqrt{3}$. We must recall our constraint from the beginning: the G.P. must be increasing.

We know $\sqrt{3}\approx 1.732$. If we choose $r=2-1.732$, we get $0.268$, which is less than $1$. This would cause our sequence to decrease, violating the problem's premise.

Therefore, we must reject the smaller root. Our only valid solution is: