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, $S_\infty$, exists if $|r| < 1$. The formula is $S_\infty = \frac{a}{1-r}$.
• GP of Squared Terms: If a GP is $a, ar, ar^2, \dots$, then the GP formed by squaring its terms is $a^2, (ar)^2, (ar^2)^2, \dots$, which simplifies to $a^2, a^2r^2, a^2r^4, \dots$. This is also an infinite GP with first term $a^2$ and common ratio $r^2$. The sum to infinity of this squared GP is $S'_\infty = \frac{a^2}{1-r^2}$, provided $|r^2| < 1$, which is true if $|r| < 1$.

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Step-by-Step Solution

Step 1: Set up the equation for the sum of the original GP. Let the first term of the given geometric progression be $a$ and the common ratio be $r$. We are given that the sum of the infinite terms of this GP is 20. Using the formula for the sum to infinity of a GP, we have: $S_\infty = \frac{a}{1-r}$ Substituting the given value: $20 = \frac{a}{1-r}$ From this, we can express the first term $a$ in terms of the common ratio $r$: $a = 20(1-r) \quad \text{(Equation 1)}$ This equation is derived directly from the problem statement and the formula for the sum of an infinite GP. We isolate $a$ to facilitate substitution later.

Step 2: Set up the equation for the sum of the squares of the GP terms. The terms of the original GP are $a, ar, ar^2, \dots$. The squares of these terms form a new sequence: $a^2, (ar)^2, (ar^2)^2, \dots$, which is $a^2, a^2r^2, a^2r^4, \dots$. This is an infinite geometric progression with:

• First term, $A = a^2$
• Common ratio, $R = r^2$

We are given that the sum of the squares of the terms to infinity is 100. Using the sum to infinity formula for this new GP: $S'_\infty = \frac{A}{1-R} = \frac{a^2}{1-r^2}$ Substituting the given value: $100 = \frac{a^2}{1-r^2}$ This equation represents the sum of the squared terms of the GP. We have identified the first term and common ratio of this new GP and applied the sum to infinity formula.

Step 3: Substitute Equation 1 into the equation from Step 2. From Equation 1, we have $a = 20(1-r)$. Squaring both sides gives us an expression for $a^2$: $a^2 = (20(1-r))^2 = 400(1-r)^2 \quad \text{(Equation 3)}$ Now, we substitute this expression for $a^2$ into the equation from Step 2 ($100 = \frac{a^2}{1-r^2}$): $100 = \frac{400(1-r)^2}{1-r^2}$ This step combines the information from both given conditions by eliminating the first term $a$. We square Equation 1 to get $a^2$ and then equate it with the expression for $a^2$ derived from the sum of squares.

Step 4: Solve the resulting equation for the common ratio 'r'. We have the equation: $100 = \frac{400(1-r)^2}{1-r^2}$ Divide both sides by 100: $1 = \frac{4(1-r)^2}{1-r^2}$ Multiply both sides by $(1-r^2)$: $1-r^2 = 4(1-r)^2$ Now, we use the difference of squares factorization for $1-r^2 = (1-r)(1+r)$: $(1-r)(1+r) = 4(1-r)^2$ For an infinite GP to converge, we must have $|r| < 1$, which implies $r \neq 1$. Therefore, $(1-r) \neq 0$, and we can divide both sides by $(1-r)$: $1+r = 4(1-r)$ Expand the right side: $1+r = 4 - 4r$ Now, we solve for $r$: Add $4r$ to both sides: $1+r+4r = 4$ $1+5r = 4$ Subtract 1 from both sides: $5r = 4-1$ $5r = 3$ Divide by 5: $r = \frac{3}{5}$ This sequence of algebraic manipulations isolates $r$. We factored $1-r^2$ and used the convergence condition $|r|<1$ to safely divide by $(1-r)$.

Step 5: Verify the solution and identify the correct option. The common ratio we found is $r = \frac{3}{5}$. We must check if this value satisfies the condition for the convergence of an infinite GP, which is $|r| < 1$. Indeed, $|\frac{3}{5}| = \frac{3}{5} < 1$. Thus, the common ratio is valid. The question asks for the common ratio of the GP. Our calculated value is $\frac{3}{5}$. Comparing this with the given options: (A) 5 (B) 3/5 (C) 8/5 (D) 1/5

Our result matches option (B).

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Common Mistakes & Tips

• Convergence Condition: Always remember that for an infinite GP sum to exist, the absolute value of the common ratio must be less than 1 ($|r|<1$). If your calculated $r$ violates this, recheck your steps.
• Sum of Squares: Be careful to correctly identify the first term ($a^2$) and the common ratio ($r^2$) for the GP formed by the squares of the original terms.
• Algebraic Simplification: When solving equations involving terms like $(1-r^2)$ and $(1-r)^2$, use factorization and the convergence condition to simplify efficiently and avoid errors.

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Summary

The problem requires the application of the sum to infinity formula for geometric progressions. We are given the sum of an infinite GP and the sum of the squares of its terms. By setting up two equations using these conditions, where the second equation involves a new GP with first term $a^2$ and common ratio $r^2$, we can eliminate the first term $a$ and solve for the common ratio $r$. The condition for convergence of an infinite GP, $|r|<1$, is crucial for simplifying the derived algebraic equation. Our calculations lead to a common ratio of $\frac{3}{5}$, which satisfies the convergence criterion.

The final answer is \boxed{3/5} which corresponds to option (B).