Problem 54 - Algebra 2

Question

What is the common ratio in a geometric series if \(a_{2}=\frac{2}{5}\) and \(a_{5}=\frac{16}{135} ?\) Enter your answer as a fraction.

Step-by-Step Solution

Answer

The common ratio 'r' is \(\frac{2\sqrt{15}}{9}\).

Step 1: Formulate the Equations

Let's express the given terms in terms of the first term and the common ratio using the formula \(a_{n} = a_{1} * r^{(n-1)}\). We will have: -\[a_{2} = a_{1} * r^{(2-1)} = a_{1} * r\]\[a_{5} = a_{1} * r^{(5-1)} = a_{1} * r^4\]We are given that \(a_{2} = \frac{2}{5}\) and \(a_{5} = \frac{16}{135}\). We substitute the values into the equations to give:- \[\frac{2}{5} = a_{1} * r\]\[\frac{16}{135} = a_{1} * r^4\]

Step 2: Solve for 'r'

To eliminate \(a_{1}\), divide the second equation by the square of the first equation to give:\[\frac{\frac{16}{135}}{(\frac{2}{5})^2} = \frac{r^4}{r^2} = r^2\]which simplifies to:\[\frac{16*25}{2*135} = r^2 \]\[\frac{200}{270} = r^2 \]\[\frac{20}{27} = r^2 \]Taking the square root of both sides, we get:\[r = \sqrt{\frac{20}{27}} = \frac{\sqrt{20}}{\sqrt{27}} = \frac{2\sqrt{5}} {3\sqrt{3}}\]

Step 3: Rationalising the Denominator

To present the answer as a whole fraction, we must rationalise the denominator. We do this by multiplying both numerator and denominator by \(\sqrt{3}\) to get:\[r = \frac{2\sqrt{5}} {3\sqrt{3}} = \frac{2\sqrt{5} * \sqrt{3}} {3\sqrt{3} * \sqrt{3}} = \frac{2\sqrt{15}} {9}\]

Key Concepts

Common Ratio

In a geometric series, the **common ratio** is the factor by which each term is multiplied to get the next term. It's denoted as \(r\). For instance, in the series \(2, 4, 8, 16, \ldots\), the common ratio is \(2\) because each term is obtained by multiplying the previous term by \(2\).
To find the common ratio, you take any term in the series and divide it by the previous term. This is helpful when you are given specific terms, such as in our example where we have \(a_2\) and \(a_5\). From these terms, we can set up an equation to solve for \(r\).
It is crucial to ensure that the given terms actually follow a geometric series pattern, otherwise the concept of a common ratio won't apply.

Algebra

Algebra involves using letters and symbols to represent numbers and quantities in equations and formulas. It allows us to express relationships and solve problems involving unknowns. In our problem, we use algebra to set up equations based on the given geometric series formula.
The equation for a term in a geometric series is \(a_n = a_1 \times r^{(n-1)}\). By substituting known values, like \(a_2\) and \(a_5\), into these equations, we can create systems of equations.

• For \(a_2\), you have \(\frac{2}{5} = a_1 \times r\).
• For \(a_5\), you have \(\frac{16}{135} = a_1 \times r^4\).

These equations allow algebraic manipulation to find unknowns by eliminating what's unnecessary, like \(a_1\), to focus solely on \(r\).

Exponential Functions

Exponential functions involve numbers raised to the power of an exponent, which, in a geometric series, is the position of the term minus one. Here, the term formula \(a_n = a_1 \times r^{(n-1)}\) demonstrates this principle.
Each position in the sequence corresponds to the exponent of \(r\). Larger exponents mean terms are further along in the series.

• For instance, \(r^1\) for the second term, or \(r^4\) for the fifth term.

This exponential nature allows the series to grow or shrink rapidly, depending on whether the common ratio \(r\) is greater than or less than one.
To determine \(r\), algebra often involves manipulating exponentials, as seen in the example where dividing equations invokes the laws of exponents to simplify the radicals.

Rationalization

Rationalization is the process of eliminating radicals, especially square roots, from the denominator in a fraction. This is a common practice in mathematics to make expressions cleaner and sometimes easier to interpret.
In our solution, the first step is to have a fraction like \(\frac{2\sqrt{5}}{3\sqrt{3}}\). The denominator includes a square root, \(\sqrt{3}\), which we want to remove.
To do this, multiply both the numerator and the denominator by \(\sqrt{3}\) to get:

• The numerator becomes \(2\sqrt{5} \times \sqrt{3} = 2\sqrt{15}\).
• The denominator becomes \(3\sqrt{3} \times \sqrt{3} = 9\), as \(\sqrt{3} \times \sqrt{3} = 3\).

Thus, the rationalized version of the fraction is \(\frac{2\sqrt{15}}{9}\). Ratios should be expressed in their simplest form, making entire fractions easier to work with in calculations or evaluations.