Problem 29 - Contemporary Precalculus

Question

The first term \(a_{1}\) and the common difference d of an arithmetic sequence are given. Find the fifth term and the formula for the nth term. $$a_{1}=10, d=-\frac{1}{2}$$

Step-by-Step Solution

Answer

Answer: The fifth term of the arithmetic sequence is 8, and the general formula is \(a_{n} = 10 + \left(n-1\right)\left(-\frac{1}{2}\right)\).

Step 1: Identify the given information in the problem

We are given the first term, \(a_{1} = 10\), and the common difference, \(d = -\frac{1}{2}\).

Step 2: Find the fifth term, \(a_{5}\)

To find the fifth term, we will use the formula \(a_{n} = a_{1} + (n-1)d\), with \(n = 5\). $$a_{5} = 10 + (5-1)\left(-\frac{1}{2}\right)$$

Step 3: Simplify the expression for \(a_{5}\)

Now, we will simplify the expression within the parentheses and then evaluate the expression: $$a_{5} = 10 + (4)\left(-\frac{1}{2}\right)$$ $$a_{5} = 10 - 4\left(\frac{1}{2}\right)$$ $$a_{5} = 10 - 2$$ $$a_{5} = 8$$ So, the fifth term of the arithmetic sequence is 8.

Step 4: Write the formula for the nth term, \(a_{n}\)

Since we already have the general formula for an arithmetic sequence, we can directly substitute the given values to obtain the formula for the given sequence: $$a_{n} = a_{1} + (n-1)d$$ $$a_{n} = 10 + (n-1)\left(-\frac{1}{2}\right)$$ Therefore, the formula for the nth term of the arithmetic sequence is: $$a_{n} = 10 + \left(n-1\right)\left(-\frac{1}{2}\right)$$

Key Concepts

Common Difference

In an arithmetic sequence, the common difference is a key feature. It tells us how much each term in the sequence increases or decreases from the previous one. Think of it as the step you take from one number to the next. For instance, if you start at 10 and your step is 2, your next number will be 12, then 14, and so on.
In simpler terms, it's the regular interval between consecutive numbers in the sequence. In our example, the common difference is given as \(d = -\frac{1}{2}\). This means that we subtract \(\frac{1}{2}\) from each term to get the next one. So, from 10, it becomes 9.5, then 9, and it keeps decreasing by \(\frac{1}{2}\).

• This consistent subtraction makes our arithmetic sequence predictable.
• It can be a positive number (sequence goes up) or negative (sequence goes down).

Remember, the common difference defines the trend of the sequence, whether it's increasing, decreasing, or staying constant.

Nth Term Formula

The nth term formula is your tool to finding any term within an arithmetic sequence, without listing all the terms! It helps you pinpoint the exact number at any position in the sequence. This formula is expressed as: \(a_{n} = a_{1} + (n-1)d\).
Here’s what each part of the formula means:

• \(a_{n}\) is the term you’re finding.
• \(a_{1}\) is the first term of the sequence.
• \(n\) is the term position you want to find.
• \(d\) is the common difference.

In our exercise, the first term, \(a_{1}\), is 10. The common difference \(d\) is \(-\frac{1}{2}\). By plugging these into the formula, we set ourselves up to find any term.
For example: To find the 5th term, you'd substitute these values into the formula and simplify. This converts a sequence problem into a simple arithmetic calculation, making it easier and faster to manage.

Sequence Term Calculation

Finding specific terms in an arithmetic sequence with the nth term formula is straightforward. Let's break down how to calculate these terms.
Using the formula \(a_{n} = a_{1} + (n-1)d\), let's see how our numbers play out:

• First, identify your first term \(a_{1}\), which is 10.
• The position \(n\) you want (for the 5th term, \(n = 5\)).
• The common difference \(d\), which is \(-\frac{1}{2}\).

Step through the calculation:
Start by replacing \(n\) with 5, and plug in your known values:
\[a_{5} = 10 + (5-1) \left(-\frac{1}{2}\right)\]
Then simplify:
\[a_{5} = 10 + 4 \left(-\frac{1}{2}\right)\]
Calculate the product inside:\[a_{5} = 10 - 2 = 8\]
This shows how you find the 5th term, which is 8. The sequence term calculation becomes easy and visual, letting you solve for any term in your sequence.