Animated Solution for Mathematics - Sequence and Series: If the sum of an infinite $GP$ $a,ar,a{r}^2,a{r}^3,\dots$ is $15$ and the sum of the squares of its each term is $150$, then the sum of $a{r}^2,a{r}^4,a{r}^6,\dots$ is :

Imagine you are standing on the edge of a mathematical precipice, looking out into the infinite. You have a sequence of numbers, $a,ar,a{r}^2,a{r}^3,\dots$, that stretches on forever.

Despite its infinite nature, it has a finite sum. This is the magic of the Geometric Progression (GP). We are given two pieces of information: the sum of the original series is $15$, and the sum of the squares of these terms is $150$.

Our goal is to find the sum of the sub-series: $a{r}^2,a{r}^4,a{r}^6,\dots$.

The sum of an infinite GP is given by the formula:

We are told this sum is $15$. Thus, we establish our first pillar of truth:

When we square every term, the new series becomes $a^2,a^2{r}^2,a^2{r}^4,\dots$. The first term is $A=a^2$ and the common ratio is $R=r^2$.

The sum of this new series is $150$. Applying the formula again, we get:

We recognize the denominator $1-r^2$ as a difference of squares, which can be factored as $(1-r)(1+r)$.

We rewrite Equation 2 using this identity:

We can split this fraction into two parts:

Since the first part is exactly Equation 1, we substitute $15$ into the expression:

Dividing both sides by $15$, we obtain:

We now have a system of two equations: $\frac{a}{1-r}=15$ and $\frac{a}{1+r}=10$. Dividing Equation 1 by Equation 3 allows the $a$ terms to cancel:

Cross-multiplying gives $2(1+r)=3(1-r)$, which simplifies to $2+2r=3-3r$. Rearranging the terms, we find $5r=1$, or:

Substituting $r=\frac{1}{5}$ back into Equation 1:

The target series $a{r}^2,a{r}^4,a{r}^6,\dots$ is an infinite GP with first term $a^{\prime }=a{r}^2$ and common ratio $r^{\prime }=r^2$. The sum is:

Substituting $a=12$ and $r=\frac{1}{5}$:

Simplifying the fraction, we arrive at the final result: