Question

If \(\mathrm{a}=1+\mathrm{r}+\mathrm{r}^{2}+\ldots \ldots \ldots \ldots \infty\) and \(|\mathrm{r}|<1\), then \(\mathrm{r}\) is (a) \(\frac{a}{a-1}\) (b) \(\frac{\mathrm{a}-1}{\mathrm{a}}\) (c) \(\frac{\mathrm{a}+1}{\mathrm{a}}\) (d) \(\frac{2 \mathrm{a}}{\mathrm{a}-1}\)

Short answer

Answer: (b) \(\frac{a - 1}{a}\)

Step-by-step solution

Identify the sum of infinite geometric series formula

Since \(a = 1 + r + r^2 + \dots + \infty\) and \(|r|<1\), we can use the formula for the sum of an infinite geometric series: \(S = \frac{A_1}{1 - r}\), where \(S\) is the sum of the series, \(A_1\) is the first term, and \(r\) is the common ratio.

Apply the formula to given series

In our case, \(S = a\), \(A_1 = 1\), and the common ratio is \(r\). So the formula becomes: \(a = \frac{1}{1 - r}\).

Solve for r

We need to solve for r. Firstly, we can multiply both sides by (1-r) to get rid of the denominator: \(a(1 - r) = 1\), Next, we distribute \(a\) on the left side: \(a - ar = 1\). Now, we isolate \(r\) on one side by moving the terms to the opposite sides: \(r = \frac{a - 1}{a}\).

Match the result with the given options

The expression we found for \(r\) is \(\frac{a - 1}{a}\), which matches option (b). Therefore, the correct answer is (b) \(\frac{a - 1}{a}\).

Key concepts

Infinite Series

When dealing with sequences of numbers, you might encounter the concept of an infinite series. This is essentially the sum of an endless list of terms. In mathematical terms, an infinite series is written as \(a + ar + ar^2 + ar^3 + \ldots \). These terms go on forever.

Infinite series become particularly interesting if the terms are getting smaller. In some cases, the sum approaches a specific number, even as more and more terms are added. This is a condition for a convergent series. The reason this happens is related to the **common ratio**, which plays a crucial part in determining the behavior of the series. If the absolute value of this ratio is less than 1, the series can converge to a finite value. In these cases, we call the series a convergent geometric series, and a special formula can be used to find the sum.

Understanding infinite series is essential in calculus and helps us to analyze phenomena like those in physics, engineering, and economics.

Common Ratio

The common ratio in a geometric series is a key player in determining the series' characteristics. This ratio is the factor by which we multiply each term to get the next term. For example, in the series \(1, r, r^2, r^3, \ldots\), **r** is the common ratio.

The intriguing part about the common ratio here is its impact on convergence. If the absolute value of the common ratio is less than one \(|r|<1\), each term gets successively smaller. Consequently, the terms approach zero, and the infinite series may sum to a finite value. This concept is essential for understanding how infinite sequences can result in a finite sum.

From an educational point of view, exploring the common ratio helps students grasp key ideas in series and sequences. It emphasizes the importance of this simple multiplier in shaping the overall behavior of an infinite series.

Sum Formula

The sum of an infinite geometric series can indeed be calculated, but only under certain conditions. The formula for finding the sum of such a series is quite elegant and straightforward:

• Sum \(S = \frac{A_1}{1 - r}\)

Here, \(A_1\) represents the first term of the series, and \(r\) is the common ratio. The beauty of this formula is its ability to distill an infinite process into a single step calculation.

When using this formula, a critical factor is the requirement that the absolute value of the common ratio must be less than one \(|r| < 1\). This ensures the terms decrease to zero, allowing the total sum to approach a specific value.

Teaching the sum formula is vital for students to accurately solve problems involving infinite series. It provides a powerful tool to evaluate these seemingly complex series, leading to solutions that are both understandable and applicable in various fields of study.