This problem involves finding the sum of an infinite geometric series of light intensities. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 10 and r = -0.6, so |r| = 0.6 < 1, ensuring convergence. The sum is S = a / (1 - r) = 10 / (1 - (-0.6)) = 10 / 1.6 = 6.25. One tempting distractor is -6.25, which might result from using 1 + r instead of 1 - r in the denominator. When working with geometric series in iterative processes like reflections, verify the sign of the ratio to apply the sum formula correctly.

This problem involves determining if an infinite geometric series converges and finding its sum if it does. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 1 and r = 5/2, so |r| = 5/2 > 1, meaning the series diverges. There is no finite sum since the terms grow without bound. One tempting distractor is -2/3, which might come from incorrectly applying the sum formula despite divergence, perhaps ignoring the condition on |r|. When evaluating geometric series, always check the convergence condition before attempting to compute the sum.

This problem involves working with geometric series to determine if the total light intensity converges. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the intensities form a series with first term 12 and r = 1.05, but |1.05| > 1, so the series diverges. The sum formula S = a / (1 - r) only applies when |r| < 1, which is not the case here. One tempting distractor is converging to 12/(1-1.05), which fails because the condition |r| < 1 is not met, making the formula invalid. A transferable strategy for geometric series is to identify the first term and common ratio, check the convergence condition |r| < 1, and only then apply the sum formula if appropriate.

This problem asks for the sum of a straightforward geometric series. The series ∑(n=0 to ∞) (3/8)(2/3)ⁿ has first term a = 3/8 (when n=0) and common ratio r = 2/3. Since |r| = 2/3 < 1, the series converges to a/(1-r) = (3/8)/(1-2/3) = (3/8)/(1/3) = (3/8)·3 = 9/8. This can also be written as (3/8)·(1/(1-2/3)) as shown in choice A. Choice B incorrectly uses 3/2 as the ratio (the reciprocal), while choice E unnecessarily includes an extra factor of 2/3. For geometric series of the form c·rⁿ starting at n=0, the sum is simply c/(1-r) when |r| < 1.

This problem tests understanding of the convergence condition for geometric series. The series ∑(n=0 to ∞) (5/2)ⁿ has first term a = 1 and common ratio r = 5/2. For a geometric series to converge, we need |r| < 1. Here, |r| = |5/2| = 5/2 = 2.5 > 1, so the series diverges. The terms (5/2)ⁿ grow without bound as n increases, preventing the series from having a finite sum. Choice A incorrectly attempts to apply the sum formula despite divergence, while choice D gives the wrong reason for divergence. Remember that geometric series converge if and only if |r| < 1; when |r| ≥ 1, the series diverges regardless of the sign of r.

This problem requires working with geometric series to find the total amount deposited in a savings account. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the deposits start with $500$ and continue with $r = 0.9$, and since $|0.9| < 1$, the series converges. The sum is $S = 500 / (1 - 0.9) = 500 / 0.1 = 5000$. One tempting distractor is 'diverges,' which fails because $|r| = 0.9 < 1$, so the series does converge to a finite value. A transferable strategy for geometric series is to identify the first term and common ratio, confirm $|r| < 1$ for convergence, and apply the sum formula for infinite terms.

This problem requires working with geometric series to find the total distance traveled by a bouncing ball. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the series is formed by the initial travel of 24 ft followed by subsequent travels scaled by r = 2/3 each time. Since |2/3| < 1, the series converges, and the sum is given by the formula S = a / (1 - r), where a = 24, yielding S = 24 / (1/3) = 72. One tempting distractor is 48, which might arise from incorrectly doubling only part of the series or misidentifying the first term. A transferable strategy for geometric series is to identify the first term and common ratio, confirm |r| < 1 for convergence, and apply the sum formula while considering the physical context.

This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the series is $4 \times \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n$, with $r = \frac{1}{3} < 1$, ensuring convergence. The sum is $$4 \times \frac{\frac{1}{3}}{1 - \frac{1}{3}} = 4 \times \frac{\frac{1}{3}}{\frac{2}{3}} = 4 \times \frac{1}{2} = 2$$. One tempting distractor is 4, which could come from mistakenly including the n=0 term or misapplying the formula. When summing geometric series starting from n=1, remember to use the formula for the tail sum appropriately.

This problem involves working with geometric series to determine if the sum of areas converges. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the areas are 9, -18, 36, -72, ..., with r = -2, but |-2| > 1, so the series diverges. The sum formula S = a / (1 - r) only applies when |r| < 1, which is not satisfied here. One tempting distractor is converging to 9/(1-(-2)), which fails because |r| > 1 violates the convergence condition, making the sum infinite. A transferable strategy for geometric series is to identify the first term and common ratio, check the convergence condition |r| < 1 first, and only apply the sum formula if the series converges.

This problem involves finding the sum of an infinite geometric series of square areas. The series is 64 + 16 + 4 + ..., where the first term a = 64 and the common ratio r = 16/64 = 1/4 = 0.25. Since |r| = 0.25 < 1, the series converges. Applying the sum formula S = a/(1-r), we get S = 64/(1-0.25) = 64/0.75 = 64/(3/4) = 256/3. Choice C (64/3) might result from forgetting the first term or making an arithmetic error. The strategy for geometric series is to identify the pattern, verify convergence with |r| < 1, then carefully apply the sum formula S = a/(1-r).