Problem 11 - College Algebra

Question

Write the first four terms of the sequence. $$a_{n}=1.25 \cdot(-4)^{n-1}$$

Step-by-Step Solution

Answer

The first four terms are 1.25, -5, 20, and -80.

Step 1: Understand the Sequence Formula

The given formula for the sequence is \(a_{n} = 1.25 \cdot (-4)^{n-1}\). This tells us that each term in the sequence is calculated by multiplying 1.25 with \((-4)\) raised to the power of \((n - 1)\), where \(n\) is the term number.

Step 2: Calculate the First Term

To find the first term \(a_1\), substitute \(n = 1\) into the formula: \[ a_1 = 1.25 \cdot (-4)^{1-1} = 1.25 \cdot (-4)^{0} = 1.25 \cdot 1 = 1.25 \] Therefore, the first term is 1.25.

Step 3: Calculate the Second Term

For the second term \(a_2\), substitute \(n = 2\) into the formula: \[ a_2 = 1.25 \cdot (-4)^{2-1} = 1.25 \cdot (-4)^{1} = 1.25 \cdot (-4) = -5 \] Thus, the second term is -5.

Step 4: Calculate the Third Term

For the third term \(a_3\), substitute \(n = 3\) into the formula: \[ a_3 = 1.25 \cdot (-4)^{3-1} = 1.25 \cdot (-4)^{2} = 1.25 \cdot 16 = 20 \] Therefore, the third term is 20.

Step 5: Calculate the Fourth Term

For the fourth term \(a_4\), substitute \(n = 4\) into the formula: \[ a_4 = 1.25 \cdot (-4)^{4-1} = 1.25 \cdot (-4)^{3} = 1.25 \cdot (-64) = -80 \] Thus, the fourth term is -80.

Key Concepts

Geometric Sequence

A geometric sequence is a series of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the sequence given by the formula \(a_{n} = 1.25 \cdot (-4)^{n-1}\), the common ratio is \(-4\). This means each term in the sequence is generated by multiplying the previous term by \(-4\).

• The sequence progresses as 1.25, -5, 20, -80, and so on.
• Notice how the sign alternates between terms; this is due to the negative common ratio.

Understanding geometric sequences helps unravel patterns found in nature, economics, and many fields of science. They are predictable and follow a structured pattern, making calculations systematic and formulated.

Term Calculation

Calculating terms in a sequence involves substituting sequential values for \(n\) into the sequence formula. For example, in \(a_{n} = 1.25 \cdot (-4)^{n-1}\), you change \(n\) to find each term:

• For the first term \(a_1\), substitute 1: \(a_1 = 1.25\cdot(-4)^0 = 1.25\).
• For the second term \(a_2\), substitute 2: \(a_2 = 1.25\cdot(-4)^1 = -5\).
• For the third term \(a_3\), substitute 3: \(a_3 = 1.25\cdot(-4)^2 = 20\).
• For the fourth term \(a_4\), substitute 4: \(a_4 = 1.25\cdot(-4)^3 = -80\).

With each substitution, a new term unveils a new facet of how geometric sequences operate. Term calculation becomes a straightforward path to understanding the progression obeyed by the sequence.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation symbols. The formula \(a_{n} = 1.25 \cdot (-4)^{n-1}\) is an algebraic expression. It combines:

• The constant 1.25
• A base, -4
• An exponent, \(n - 1\)

When dealing with algebraic expressions, understanding how each part interacts is crucial.

• The base \(-4\) and its exponent determine the variable part's growth in the expression.
• The multiplier 1.25 scales the changing values to fit the sequence's specific progression.

Recognizing these components empowers you to manipulate and solve sequence-related tasks more smoothly with algebraic reasoning.

Exponentiation

Exponentiation is the operation of raising one number, the base, to the power of another, the exponent. Within our sequence, exponentiation governs the sequence progression as every term is calculated by raising \(-4\) to increasing powers (\(n - 1\)).

• The zero exponent rule states \((-4)^0 = 1\), yielding the first term as 1.25 multiplied by 1.
• For the second term, the exponent \(1\) means the base remains unchanged: \( (-4)^1 = -4 \).
• With the third term, the exponent is \(2\), thus \( (-4)^2 = 16 \), resulting in a positive product.
• The pattern switches back to negative as \( (-4)^3 = -64 \).

Understand that the sign depends on whether the exponent is odd (negative result) or even (positive result). This property is critical in managing sequences that alternate in sign, as in this example.