Problem 21 - Precalculus: Mathematical for Calculus

Question

Find the first five terms of the sequence and determine if it is arithmetic. If it is arithmetic, find the common difference and express the \(n\) th term of the sequence in the standard form $$a_{n}=a+(n-1) d$$ $$a_{n}=6 n-10$$

Step-by-Step Solution

Answer

The sequence is arithmetic with a common difference of 6, and the expression for \(a_{n}\) is already given as \(6n - 10\).

Step 1: Substitute to Find Terms

Substitute the values of \(n\) (1 through 5) into the formula \(a_{n} = 6n - 10\) to find the first five terms of the sequence. This follows the order:- For \(n=1\), \(a_1 = 6(1)-10 = -4\).- For \(n=2\), \(a_2 = 6(2)-10 = 2\).- For \(n=3\), \(a_3 = 6(3)-10 = 8\).- For \(n=4\), \(a_4 = 6(4)-10 = 14\).- For \(n=5\), \(a_5 = 6(5)-10 = 20\).

Step 2: Assess Arithmetic Sequence

To determine if the sequence is arithmetic, check if the difference between consecutive terms is constant. Calculate the differences:- \(a_2 - a_1 = 2 - (-4) = 6\).- \(a_3 - a_2 = 8 - 2 = 6\).- \(a_4 - a_3 = 14 - 8 = 6\).- \(a_5 - a_4 = 20 - 14 = 6\).The difference is constant at 6, confirming the sequence is arithmetic.

Step 3: Identify Common Difference and Express Formula

Since the sequence is arithmetic and the common difference \(d\) is 6, express the \(n\)th term using the standard formula for an arithmetic sequence:\[a_{n} = a + (n-1) d\]Here, \(a = a_1 = -4\) and \(d = 6\). Substituting these values gives:\[a_{n} = -4 + (n-1) \cdot 6\]Simplify it to:\[a_{n} = -4 + 6n - 6\]\[a_{n} = 6n - 10\]The formula matches the original expression, confirming the calculations.

Key Concepts

Common Difference

In an arithmetic sequence, understanding the concept of the common difference is key. The common difference is the consistent amount between consecutive terms in the sequence. This difference remains the same throughout the sequence, providing it with a uniform structure.
Here's how to find it: subtract any term from the subsequent term. If the difference does not change as you move from one term to the next, the sequence is indeed arithmetic.
For the given sequence, we calculate:

• Difference between second term and first: \( a_2 - a_1 = 2 - (-4) = 6 \).
• Difference between third term and second: \( a_3 - a_2 = 8 - 2 = 6 \).
• Continue the pattern for other terms

These equal differences confirm our sequence has a common difference of 6, ensuring it fits the criteria of an arithmetic sequence.

Nth Term Formula

To express the general term of an arithmetic sequence, we use the nth term formula. This captures the consistent pattern of an arithmetic sequence in a simple mathematical expression.
The formula is given by:\[ a_n = a + (n-1) \cdot d \] where

• \( a \) is the first term of the sequence.
• \( d \) is the common difference.
• \( n \) is the term number in the sequence.

For the example sequence, starting from \( a_1 = -4 \) and \( d = 6 \), substituting into the formula gives:

• \( a_n = -4 + (n-1) \cdot 6 \)

Simplifying it confirms \( a_n = 6n - 10 \), matching the original nth term function. This formula helps easily find any term in the sequence without computing every proceeding term.

Sequence Terms

The sequence terms represent individual numbers in a sequence, each corresponding to a particular position or term number \( n \). For clarity, let's find the first few terms in the sequence given by the formula \( a_n = 6n - 10 \).
Calculating manually:

• For \( n=1 \): \( a_1 = 6(1) - 10 = -4 \)
• For \( n=2 \): \( a_2 = 6(2) - 10 = 2 \)
• For \( n=3 \): \( a_3 = 6(3) - 10 = 8 \)
• For \( n=4 \): \( a_4 = 6(4) - 10 = 14 \)
• For \( n=5 \): \( a_5 = 6(5) - 10 = 20 \)

These operations help visualize how the sequence develops, making it easier to see the arithmetic pattern established by the formula.