Question:

Step 1: Let the geometric progression be $a, ar, ar^2, \ldots$. For the sum of an infinite geometric progression to converge, the common ratio $r$ must satisfy $|r| < 1$. The sum of the infinite geometric progression is given by: $$S = \frac{a}{1-r} = 20 \quad (1)$$ Step 2: The terms of the series of squares are $a^2, a^2r^2, a^2r^4, \ldots$. This is also an infinite geometric progression with first term $a^2$ and common ratio $r^2$. Since $|r|<1$, it follows that $|r^2|<1$. The sum of the squares is given by: $$S' = \frac{a^2}{1-r^2} = 100 \quad (2)$$ Step 3: From equation (1), express $a$ in terms of $r$: $$a = 20(1-r)$$ Step 4: Substitute this expression for $a$ into equation (2): $$\frac{[20(1-r)]^2}{1-r^2} = 100$$ Step 5: Simplify the equation. $$\frac{400(1-r)^2}{(1-r)(1+r)} = 100$$ Divide both sides by 100: $$\frac{4(1-r)^2}{(1-r)(1+r)} = 1$$ Multiply both sides by $(1-r)(1+r)$: $$4(1-r)^2 = (1-r)(1+r)$$ $$4(1-r)^2 = 1-r^2$$ Step 6: Expand and rearrange the equation into a quadratic form: $$4(1 - 2r + r^2) = 1 - r^2$$ $$4 - 8r + 4r^2 = 1 - r^2$$ $$5r^2 - 8r + 3 = 0$$ Step 7: Solve the quadratic equation for $r$. Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $$r = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(3)}}{2(5)}$$ $$r = \frac{8 \pm \sqrt{64 - 60}}{10}$$ $$r = \frac{8 \pm \sqrt{4}}{10}$$ $$r = \frac{8 \pm 2}{10}$$ This yields two possible values for $r$: $$r_1 = \frac{8 + 2}{10} = \frac{10}{10} = 1$$ $$r_2 = \frac{8 - 2}{10} = \frac{6}{10} = \frac{3}{5}$$ Step 8: Select the valid common ratio. For an infinite geometric series to converge, the common ratio $r$ must satisfy the condition $|r| < 1$. The value $r=1$ does not satisfy this condition, as it would lead to a divergent series. Therefore, the valid common ratio is $r = \frac{3}{5}$. Step 9: Verify the solution. Using $r = 3/5$, we find the first term $a$ from equation (1): $$a = 20(1-r) = 20(1 - 3/5) = 20(2/5) = 8$$ Now, check if these values satisfy the condition for the sum of squares (equation 2): $$S' = \frac{a^2}{1-r^2} = \frac{8^2}{1-(3/5)^2} = \frac{64}{1-9/25} = \frac{64}{(25-9)/25} = \frac{64}{16/25} = 64 \cdot \frac{25}{16} = 4 \cdot 25 = 100$$ The calculated sum of squares matches the given value of 100. Thus, the common ratio of the G.P. is $\frac{3}{5}$.