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Question 1
An infinite geometric series has first term $u_{1} = 7$ and common ratio $r$ . Which of the following values of $r$ ensures the series has a finite sum to infinity?
No clue? Show me the answer
Correct answerCorrect!Incorrect
C $r = - \frac{5}{7}$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: State the convergence condition
An infinite geometric series converges if and only if $∣ r ∣ < 1$ . We need to find which option satisfies this condition.
Step 2: Test each option
Check $- \frac{4}{3} = \frac{4}{3} > 1$ (diverges), $∣1∣ = 1 \geq 1$ (diverges), $- \frac{5}{7} = \frac{5}{7} < 1$ (converges), $\frac{7}{5} = \frac{7}{5} > 1$ (diverges).
Step 3: Identify the convergent case
Only $r = - \frac{5}{7}$ satisfies $∣ r ∣ < 1$ , so this is the only value that guarantees a finite sum $S_{\infty}$ .
Method #2Approach 2
Step 1: What is being asked
We need the value of $r$ for which $S_{\infty} = \frac{u _{1}}{1 - r}$ exists and is finite, requiring $∣ r ∣ < 1$ .
Step 2: Eliminate $r = -\dfrac{4}{3}$
$- \frac{4}{3} = \frac{4}{3} > 1$ , so the terms grow in magnitude and the series diverges. Eliminated.
Step 3: Eliminate $r = 1$
When $r = 1$ , every term equals $u_{1} = 7$ , so the partial sums grow without bound. The series diverges. Eliminated.
Step 4: Eliminate $r = \dfrac{7}{5}$
$\frac{7}{5} = 1.4 > 1$ , so terms grow geometrically and the series diverges. Eliminated.
Step 5: Select $r = -\dfrac{5}{7}$
$- \frac{5}{7} = \frac{5}{7} \approx 0.714 < 1$ , satisfying the convergence condition. This is the correct answer.
Question 2
The concentration of a drug in a patient's bloodstream (in mg/L) at the end of each hour is modelled by $C_{n} = 80 (\frac{3}{4})^{n}$ , where $n \geq 1$ . If the concentrations at each hour are summed indefinitely, what is the value of $n = 1 \sum \infty C_{n}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $240$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the series structure
The sum is $n = 1 \sum \infty 80 (\frac{3}{4})^{n}$ . The first term (at $n = 1$ ) is $u_{1} = 80 \times \frac{3}{4} = 60$ , and the common ratio is $r = \frac{3}{4}$ .
Step 2: Verify convergence
$∣ r ∣ = \frac{3}{4} < 1$ , so the series converges and $S_{\infty}$ exists.
Step 3: Apply the sum to infinity formula
$S_{\infty} = \frac{u _{1}}{1 - r} = \frac{60}{1 - \frac{3}{4}} = \frac{60}{\frac{1}{4}} = 60 \times 4 = 240$
Step 4: State the answer
The infinite sum of drug concentrations is $240$ mg/L.
Method #2Approach 2
Step 1: Identify first term and ratio
At $n = 1$ : $C_{1} = 80 \times \frac{3}{4} = 60$ . The ratio between consecutive terms is $r = \frac{3}{4}$ , so $S_{\infty} = \frac{60}{1 - 3/4} = 240$ .
Step 2: Eliminate $320$
$320$ would result from using $u_{1} = 80$ (the $n = 0$ term), but since $n \geq 1$ , the first term in this sum is $60$ , not $80$ . Eliminated.
Step 3: Eliminate $60$
$60$ is just the first term $C_{1}$ , not the infinite sum. The sum of infinitely many positive terms must exceed any single term. Eliminated.
Step 4: Eliminate $180$
$180$ might arise from incorrectly computing $\frac{60}{1 - 3/4} = \frac{60}{1/4}$ as $60 \times 3 = 180$ instead of $60 \times 4 = 240$ . This is an arithmetic error. Eliminated.
Step 5: Select $240$
$S_{\infty} = \frac{60}{\frac{1}{4}} = 240$ is correct. The answer is $240$ .

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Question 1
An infinite geometric series has first term $u_{1} = 7$ and common ratio $r$ . Which of the following values of $r$ ensures the series has a finite sum to infinity?
No clue? Show me the answer
Correct answerCorrect!Incorrect
C $r = - \frac{5}{7}$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: State the convergence condition
An infinite geometric series converges if and only if $∣ r ∣ < 1$ . We need to find which option satisfies this condition.
Step 2: Test each option
Check $- \frac{4}{3} = \frac{4}{3} > 1$ (diverges), $∣1∣ = 1 \geq 1$ (diverges), $- \frac{5}{7} = \frac{5}{7} < 1$ (converges), $\frac{7}{5} = \frac{7}{5} > 1$ (diverges).
Step 3: Identify the convergent case
Only $r = - \frac{5}{7}$ satisfies $∣ r ∣ < 1$ , so this is the only value that guarantees a finite sum $S_{\infty}$ .
Method #2Approach 2
Step 1: What is being asked
We need the value of $r$ for which $S_{\infty} = \frac{u _{1}}{1 - r}$ exists and is finite, requiring $∣ r ∣ < 1$ .
Step 2: Eliminate $r = -\dfrac{4}{3}$
$- \frac{4}{3} = \frac{4}{3} > 1$ , so the terms grow in magnitude and the series diverges. Eliminated.
Step 3: Eliminate $r = 1$
When $r = 1$ , every term equals $u_{1} = 7$ , so the partial sums grow without bound. The series diverges. Eliminated.
Step 4: Eliminate $r = \dfrac{7}{5}$
$\frac{7}{5} = 1.4 > 1$ , so terms grow geometrically and the series diverges. Eliminated.
Step 5: Select $r = -\dfrac{5}{7}$
$- \frac{5}{7} = \frac{5}{7} \approx 0.714 < 1$ , satisfying the convergence condition. This is the correct answer.
Question 2
The concentration of a drug in a patient's bloodstream (in mg/L) at the end of each hour is modelled by $C_{n} = 80 (\frac{3}{4})^{n}$ , where $n \geq 1$ . If the concentrations at each hour are summed indefinitely, what is the value of $n = 1 \sum \infty C_{n}$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $240$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the series structure
The sum is $n = 1 \sum \infty 80 (\frac{3}{4})^{n}$ . The first term (at $n = 1$ ) is $u_{1} = 80 \times \frac{3}{4} = 60$ , and the common ratio is $r = \frac{3}{4}$ .
Step 2: Verify convergence
$∣ r ∣ = \frac{3}{4} < 1$ , so the series converges and $S_{\infty}$ exists.
Step 3: Apply the sum to infinity formula
$S_{\infty} = \frac{u _{1}}{1 - r} = \frac{60}{1 - \frac{3}{4}} = \frac{60}{\frac{1}{4}} = 60 \times 4 = 240$
Step 4: State the answer
The infinite sum of drug concentrations is $240$ mg/L.
Method #2Approach 2
Step 1: Identify first term and ratio
At $n = 1$ : $C_{1} = 80 \times \frac{3}{4} = 60$ . The ratio between consecutive terms is $r = \frac{3}{4}$ , so $S_{\infty} = \frac{60}{1 - 3/4} = 240$ .
Step 2: Eliminate $320$
$320$ would result from using $u_{1} = 80$ (the $n = 0$ term), but since $n \geq 1$ , the first term in this sum is $60$ , not $80$ . Eliminated.
Step 3: Eliminate $60$
$60$ is just the first term $C_{1}$ , not the infinite sum. The sum of infinitely many positive terms must exceed any single term. Eliminated.
Step 4: Eliminate $180$
$180$ might arise from incorrectly computing $\frac{60}{1 - 3/4} = \frac{60}{1/4}$ as $60 \times 3 = 180$ instead of $60 \times 4 = 240$ . This is an arithmetic error. Eliminated.
Step 5: Select $240$
$S_{\infty} = \frac{60}{\frac{1}{4}} = 240$ is correct. The answer is $240$ .

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Step-by-step walkthrough

Step 1: State the convergence condition

Step 2: Test each option

Step 3: Identify the convergent case

Step 1: What is being asked

Step 2: Eliminate $r = -\dfrac{4}{3}$

Step 3: Eliminate $r = 1$

Step 4: Eliminate $r = \dfrac{7}{5}$

Step 5: Select $r = -\dfrac{5}{7}$

Step-by-step walkthrough

Step 1: Identify the series structure

Step 2: Verify convergence

Step 3: Apply the sum to infinity formula

Step 4: State the answer

Step 1: Identify first term and ratio

Step 2: Eliminate $320$

Step 3: Eliminate $60$

Step 4: Eliminate $180$

Step 5: Select $240$

14 more questions in this topic

Step-by-step walkthrough

Step 1: State the convergence condition

Step 2: Test each option

Step 3: Identify the convergent case

Step 1: What is being asked

Step 2: Eliminate $r = -\dfrac{4}{3}$

Step 3: Eliminate $r = 1$

Step 4: Eliminate $r = \dfrac{7}{5}$

Step 5: Select $r = -\dfrac{5}{7}$

Step-by-step walkthrough

Step 1: Identify the series structure

Step 2: Verify convergence

Step 3: Apply the sum to infinity formula

Step 4: State the answer

Step 1: Identify first term and ratio

Step 2: Eliminate $320$

Step 3: Eliminate $60$

Step 4: Eliminate $180$

Step 5: Select $240$

14 more questions in this topic