AP Calculus BC Practice Quiz: Working with Geometric Series

Correct Answer: A

This is a geometric series of the form $\sum_{n=0}^{\infty} a r^n$ with the first term $a=4$ and the common ratio $r = \frac{1}{3}$. Since $|r| = |\frac{1}{3}| < 1$, the series converges. The sum is given by the formula $\frac{a}{1-r} = \frac{4}{1 - \frac{1}{3}} = \frac{4}{\frac{2}{3}} = 4 \cdot \frac{3}{2} = 6$.

Correct Answer: D

A geometric series $\sum_{n=0}^{\infty} a r^n$ converges only if the absolute value of the common ratio, $|r|$, is less than 1. In this series, the common ratio is $r = \frac{5}{4}$. Since $|r| = |\frac{5}{4}| > 1$, the series diverges.

Correct Answer: B

The series can be rewritten to fit the standard form of a geometric series, $\sum_{n=0}^{\infty} a r^n$. We have $\sum_{n=0}^{\infty} \frac{3^{n+1}}{5^n} = \sum_{n=0}^{\infty} \frac{3 \cdot 3^n}{5^n} = \sum_{n=0}^{\infty} 3 \left(\frac{3}{5}\right)^n$. Here, the first term is $a=3$ and the common ratio is $r = \frac{3}{5}$. Since $|r| < 1$, the series converges to $\frac{a}{1-r} = \frac{3}{1 - \frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}$.

Correct Answer: A

This is a geometric series with first term $a=2$ and common ratio $r = -\frac{1}{2}$. Since $|r| = |-\frac{1}{2}| = \frac{1}{2} < 1$, the series converges. The sum is $\frac{a}{1-r} = \frac{2}{1 - (-\frac{1}{2})} = \frac{2}{1 + \frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$.

Correct Answer: C

According to the convergence test for a geometric series, the series $\sum_{n=0}^{\infty} a r^n$ converges if and only if the absolute value of the common ratio $r$ is less than 1, which is expressed as $|r| < 1$.

Correct Answer: B

This is a geometric series with a common ratio $r = \frac{1}{4}$. Since the series starts at $n=1$, the first term is $a = 5(\frac{1}{4})^1 = \frac{5}{4}$. Because $|r| < 1$, the series converges. The sum of a geometric series is the first term divided by (1 minus the common ratio), so the sum is $\frac{a}{1-r} = \frac{\frac{5}{4}}{1 - \frac{1}{4}} = \frac{\frac{5}{4}}{\frac{3}{4}} = \frac{5}{3}$.

Correct Answer: C

A geometric series diverges if its common ratio $r$ has an absolute value greater than or equal to 1. Let's analyze the options: A) $r = 1/2$, $|r| < 1$, converges. B) $r = 0.99$, $|r| < 1$, converges. C) The series can be written as $\sum_{n=0}^{\infty} (-\frac{3}{2})^n$. Here, $r = -3/2$, and $|r| = 3/2 > 1$, so it diverges. D) $r = -4/5$, $|r| < 1$, converges.

Correct Answer: C

The sum of a convergent geometric series is given by the formula $S = \frac{a}{1-r}$. We are given $S=15$ and $a=5$. Plugging these values into the formula gives $15 = \frac{5}{1-r}$. To solve for $r$, we can multiply both sides by $(1-r)$ to get $15(1-r) = 5$. Distributing gives $15 - 15r = 5$. Subtracting 15 from both sides gives $-15r = -10$. Finally, dividing by -15 gives $r = \frac{-10}{-15} = \frac{2}{3}$.

Correct Answer: C

The repeating decimal can be expressed as the series $0.27 + 0.0027 + 0.000027 + \dots$. This is a geometric series with the first term $a = 0.27 = \frac{27}{100}$ and a common ratio $r = \frac{0.0027}{0.27} = \frac{1}{100}$. Since $|r| = \frac{1}{100} < 1$, the series converges. The sum is $\frac{a}{1-r} = \frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{\frac{27}{100}}{\frac{99}{100}} = \frac{27}{99}$. Simplifying this fraction by dividing the numerator and denominator by 9 gives $\frac{3}{11}$.