Problem 55 - College Algebra

Question

At which term does the sequence \(\\{10,12,14.4,17.28, \ldots\\}\) exceed \(100 ?\)

Step-by-Step Solution

Answer

The term exceeds 100 at the 14th position.

Step 1: Identify the Sequence Type

First, observe the sequence \(\{10, 12, 14.4, 17.28, \ldots\}\). Notice how each term is obtained by multiplying the previous term by a constant factor, making it a geometric sequence.

Step 2: Determine the Common Ratio

In a geometric sequence, each term is found by multiplying the previous term by the common ratio \(r\). Calculate \(r\) by dividing the second term by the first term: \(r = \frac{12}{10} = 1.2\).

Step 3: Establish the General Formula

For a geometric sequence, the \(n\)-th term \(a_n\) can be represented as \(a_n = a_1 \cdot r^{n-1}\), where \(a_1\) is the first term. Here, \(a_1 = 10\), so the formula becomes \(a_n = 10 \cdot (1.2)^{n-1}\).

Step 4: Solve for \(n\) When \(a_n > 100\)

We need to find the smallest \(n\) such that \(a_n = 10 \cdot (1.2)^{n-1} > 100\). We set up the inequality:\[10 \cdot (1.2)^{n-1} > 100\]Divide both sides by 10:\[(1.2)^{n-1} > 10\]

Step 5: Use Logarithms to Solve the Inequality

Take the logarithm of both sides to solve for \(n\):\[\log((1.2)^{n-1}) > \log(10)\]This simplifies to:\[(n-1) \cdot \log(1.2) > 1\]Divide both sides by \(\log(1.2)\):\[n-1 > \frac{1}{\log(1.2)}\]

Step 6: Calculate \(n\)

Calculate \(n\) by solving \[n > \frac{1}{\log(1.2)} + 1\].Using a calculator, \(\log(1.2) \approx 0.07918\).\[n > \frac{1}{0.07918} + 1 \approx 13.63\]Thus the smallest integer \(n\) is 14.

Key Concepts

Common Ratio

In a geometric sequence, each term after the first is derived by multiplying the previous term by the common ratio. Understanding the common ratio is key to predicting the sequence's progression. For the sequence \(\{10, 12, 14.4, 17.28, \ldots\}\), we calculate the common ratio \(r\) by dividing the second term by the first term: \(r = \frac{12}{10} = 1.2\).
This simple calculation reveals a lot about the sequence. It's an indication of how the sequence grows. In this case, because the common ratio is greater than 1 (\(1.2\)), the terms in the sequence will grow exponentially.
Remember, the common ratio could also be less than 1, indicating a sequence that decreases, or negative, leading to alternating terms. But here, 1.2 signifies consistent growth.

nth Term

The \(n\)-th term in a geometric sequence gives us flexibility for determining any term in the sequence without listing all previous terms. This helps greatly with making predictions or solving for specific sequence benchmarks.
The general formula for the \(n\)-th term \(a_n\) of a geometric sequence is \(a_n = a_1 \cdot r^{n-1}\), where \(a_1\) is the first term and \(r\) is the common ratio. For our sequence, this becomes \(a_n = 10 \cdot (1.2)^{n-1}\).
This formula is crucial as it allows us to set up equations and inequalities to find specific terms or the point at which a certain condition is met, like exceeding 100 in this sequence.

Logarithms

Logarithms are vital for working backwards from exponential equations, like those found in geometric sequences. They transform multiplicative processes into additive ones, simplifying calculations significantly.
In this example, we use logarithms to solve the inequality \((1.2)^{n-1} > 10\). By applying logarithms, we get \(\log((1.2)^{n-1}) > \log(10)\), which simplifies to \((n-1) \cdot \log(1.2) > 1\).
This transformation makes it easier to isolate \(n\) and ultimately solve for it. Logarithms are powerful tools, particularly when dealing with exponential growth or decay situations, and they help us find solutions that would otherwise be challenging to solve manually.

Inequality Solving

Solving inequalities, particularly in geometric sequences, sometimes requires some extra steps compared to simple equalities. We are solving to find when a specific condition is met—in our problem, when the term exceeds 100.
Once we've set up the inequality \((1.2)^{n-1} > 10\), we use logarithms to simplify it to \(n-1 > \frac{1}{\log(1.2)}\). By rearranging, we get \(n > 13.63\), meaning \(n\) needs to be the next whole number since it represents a term count.
Ultimately, solving this inequality determined that the sequence's 14th term is the first to exceed 100. Breaking down the problem this way clarifies an often challenging concept, reinforcing the strong utility of both logarithms and inequalities in solving exponential growth questions.