Key Concepts and Formulas

• Sum of an Infinite Geometric Progression (GP): For a GP with first term $A$ and common ratio $R$, the sum to infinity is $S = \frac{A}{1-R}$, provided $|R| < 1$.
• Properties of Exponents: $(x^m)^n = x^{mn}$.
• Algebraic Identity: $1-x^2 = (1-x)(1+x)$.

Step-by-Step Solution

Step 1: Setting up equations from the given information

We are given an infinite GP: $a, ar, ar^2, ar^3, \ldots$. The sum of this GP is 15. Using the formula for the sum of an infinite GP, we have: $\frac{a}{1-r} = 15 \quad \ldots (1)$ This implies $|r| < 1$.

The series formed by the squares of each term is $a^2, (ar)^2, (ar^2)^2, (ar^3)^2, \ldots$, which simplifies to $a^2, a^2r^2, a^2r^4, a^2r^6, \ldots$. This is also an infinite GP with the first term $A' = a^2$ and the common ratio $R' = r^2$. Since $|r| < 1$, we have $|r^2| < 1$, so the sum of this series also exists. The sum of the squares is given as 150: $\frac{a^2}{1-r^2} = 150 \quad \ldots (2)$

Step 2: Simplifying Equation (2) and relating it to Equation (1)

We can rewrite Equation (2) using the identity $1-r^2 = (1-r)(1+r)$: $\frac{a^2}{(1-r)(1+r)} = 150$ This can be split as: $\left(\frac{a}{1-r}\right) \cdot \left(\frac{a}{1+r}\right) = 150$ Substitute the value of $\frac{a}{1-r}$ from Equation (1) into this expression: $15 \cdot \left(\frac{a}{1+r}\right) = 150$ Dividing by 15, we get: $\frac{a}{1+r} = 10 \quad \ldots (3)$

Step 3: Solving for the common ratio ($r$) and the first term ($a$)

We now have a system of two equations:

1. $\frac{a}{1-r} = 15$
2. $\frac{a}{1+r} = 10$

Divide Equation (1) by Equation (3): $\frac{\frac{a}{1-r}}{\frac{a}{1+r}} = \frac{15}{10}$ $\frac{a}{1-r} \cdot \frac{1+r}{a} = \frac{3}{2}$ $\frac{1+r}{1-r} = \frac{3}{2}$ Cross-multiply to solve for $r$: $2(1+r) = 3(1-r)$ $2 + 2r = 3 - 3r$ $5r = 1$ $r = \frac{1}{5}$ Since $r = \frac{1}{5}$ and $|r| < 1$, this is a valid common ratio.

Now, substitute $r = \frac{1}{5}$ into Equation (3) to find $a$: $\frac{a}{1+\frac{1}{5}} = 10$ $\frac{a}{\frac{6}{5}} = 10$ $a = 10 \cdot \frac{6}{5} = 12$ So, the first term is $a=12$ and the common ratio is $r=\frac{1}{5}$.

Step 4: Identifying the target series and calculating its sum

We need to find the sum of the series $ar^2, ar^4, ar^6, \ldots$. This is an infinite GP with:

• First term: $A'' = ar^2$
• Common ratio: $R'' = \frac{ar^4}{ar^2} = r^2$

We have found $a=12$ and $r=\frac{1}{5}$. So, the first term of this new series is: $A'' = ar^2 = 12 \cdot \left(\frac{1}{5}\right)^2 = 12 \cdot \frac{1}{25} = \frac{12}{25}$ The common ratio of this new series is: $R'' = r^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25}$ Since $|R''| = \frac{1}{25} < 1$, the sum of this infinite GP exists. The sum is: $S_{\text{target}} = \frac{A''}{1-R''} = \frac{\frac{12}{25}}{1-\frac{1}{25}}$ $S_{\text{target}} = \frac{\frac{12}{25}}{\frac{25-1}{25}} = \frac{\frac{12}{25}}{\frac{24}{25}}$ $S_{\text{target}} = \frac{12}{25} \cdot \frac{25}{24} = \frac{12}{24} = \frac{1}{2}$

Common Mistakes & Tips

• Condition for Infinite Sum: Always remember that the sum of an infinite GP converges only if the absolute value of the common ratio is strictly less than 1 (i.e., $|r| < 1$).
• Common Ratio of Squared Series: Be careful to correctly identify the common ratio of the series of squares. If the original common ratio is $r$, the common ratio of the squared series is $r^2$.
• Algebraic Manipulation: Practice simplifying expressions involving fractions and algebraic identities to avoid calculation errors.

Summary

The problem involves finding the sum of a specific infinite geometric progression given the sum of the original GP and the sum of the squares of its terms. We first set up equations based on the given information and the formula for the sum of an infinite GP. By manipulating these equations, we solved for the first term ($a$) and the common ratio ($r$) of the original GP. Finally, we used these values to determine the first term and common ratio of the target series ($ar^2, ar^4, ar^6, \ldots$) and calculated its sum.

The final answer is \boxed{${1 \over 2}$}.