Finding The Linear Function Of An Arithmetic Sequence

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Hey math enthusiasts! Today, we're diving into the fascinating world of arithmetic sequences and linear functions. Let's break down how to find the linear function given some specific terms of an arithmetic sequence. We'll solve a problem where the third term is 34\frac{3}{4} and the ninth term is 154\frac{15}{4}. Get ready to explore the relationship between arithmetic sequences and linear equations!

Understanding Arithmetic Sequences

Alright, guys, before we jump into the problem, let's make sure we're all on the same page about arithmetic sequences. An arithmetic sequence is simply a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, often denoted by 'd'. Think of it like this: you start with a number, and you repeatedly add the same amount to get the next number in the sequence. For example, the sequence 2, 5, 8, 11... is an arithmetic sequence with a common difference of 3. Each term is obtained by adding 3 to the previous term.

Now, how does this relate to linear functions? Well, each arithmetic sequence can be represented by a linear function. The terms of the sequence correspond to points on the line. The common difference of the sequence is the slope of the line, and the first term helps us find the y-intercept. So, when we're dealing with an arithmetic sequence, we're essentially dealing with a straight line. Pretty cool, huh? The general form of an arithmetic sequence is an=a1+(nβˆ’1)da_n = a_1 + (n-1)d, where ana_n is the nth term, a1a_1 is the first term, nn is the term number, and dd is the common difference. This is super important!

To really get this, let's visualize a simple example. Suppose we have an arithmetic sequence 1, 3, 5, 7.... The first term, a1a_1, is 1, and the common difference, dd, is 2. The formula gives us an=1+(nβˆ’1)2a_n = 1 + (n-1)2. If we plot these points (1,1), (2,3), (3,5), (4,7) on a graph, we will see a straight line. This line represents the linear function associated with the sequence. Understanding this fundamental link between arithmetic sequences and linear functions is key to solving our problem.

Setting up the Problem

Okay, let's get down to the nitty-gritty. The problem gives us two pieces of vital info: the third term (a3a_3) of the arithmetic sequence is 34\frac{3}{4}, and the ninth term (a9a_9) is 154\frac{15}{4}. Our goal is to find the linear function that represents this sequence. We can write the general formula for any term as f(x)=a+(xβˆ’1)df(x) = a + (x-1)d where aa is the first term and dd is the common difference, which is like the slope in a linear equation. We can think of the position of the term as the x value. So let's start by noting down the known information. We know that when x = 3, f(x) = 34\frac{3}{4} and when x = 9, f(x) = 154\frac{15}{4}.

We need to find this linear function f(x)f(x). Since we're dealing with a linear function, we know it can be written in the form f(x)=mx+bf(x) = mx + b, where 'm' is the slope (the common difference 'd' in our arithmetic sequence) and 'b' is the y-intercept. But we are also told about two specific points. Using these points, we are going to find the slope (common difference) first, and then find the y intercept by solving a system of linear equations. This will help us completely define the linear function that describes our arithmetic sequence. The common difference 'd' remains constant throughout the sequence. The key is to see how the change in the term number impacts the change in the term values. Let's get to work!

Finding the Common Difference (Slope)

Alright, folks, time to find that all-important common difference, 'd'. We have two terms and their positions in the sequence. Remember that the common difference is constant. We can find this by figuring out how much the terms change over the change in the position. We can set up the slope equation using our given points (3, 34\frac{3}{4}) and (9, 154\frac{15}{4}). The slope formula is: m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}. In our case, x represents the term number and y represents the value of the term. Let's plug in our values and see what we get:

d=154βˆ’349βˆ’3d = \frac{\frac{15}{4} - \frac{3}{4}}{9 - 3}

Now, let's simplify this. First, subtract the fractions in the numerator. 154βˆ’34=124=3\frac{15}{4} - \frac{3}{4} = \frac{12}{4} = 3. Then, calculate the difference in the denominator: 9βˆ’3=69 - 3 = 6. So, our equation becomes:

d=36=12d = \frac{3}{6} = \frac{1}{2}

Awesome! We've found the common difference, d=12d = \frac{1}{2}. This is also the slope of our linear function. Now that we know the slope, our linear function looks like f(x)=12x+bf(x) = \frac{1}{2}x + b. We are now on the right track! The next step is to find the y-intercept. We'll use one of the given points and the slope we just calculated to determine the value of 'b', the y-intercept. Knowing both slope and y-intercept allows us to precisely define the linear function. Let's move on!

Determining the Linear Function

Now we're in the final stretch, guys! We've got the common difference (the slope), d=12d = \frac{1}{2}. Remember, our linear function looks like f(x)=12x+bf(x) = \frac{1}{2}x + b. We can plug in one of the points we know from the problem to solve for 'b'. Let's use the point where x=3x=3 and f(x)=34f(x) = \frac{3}{4}. So, we have:

34=12(3)+b\frac{3}{4} = \frac{1}{2}(3) + b

Simplify the right side: 12(3)=32\frac{1}{2}(3) = \frac{3}{2}. Now our equation becomes:

34=32+b\frac{3}{4} = \frac{3}{2} + b

To isolate 'b', we subtract 32\frac{3}{2} from both sides. But we need a common denominator to make the subtraction easy. So let's convert 32\frac{3}{2} to 64\frac{6}{4}. Now we have:

34βˆ’64=b\frac{3}{4} - \frac{6}{4} = b

βˆ’34=b\frac{-3}{4} = b

So, the y-intercept, b, is βˆ’34\frac{-3}{4}. We now have everything we need to write the complete linear function. Plugging our values into the function we get:

f(x)=12xβˆ’34f(x) = \frac{1}{2}x - \frac{3}{4}

But wait! The multiple choice answers aren't in this format. We need to manipulate our function into the format of the options. This can be done by rearranging the formula from a point-slope form. We know our point is (3, 34\frac{3}{4}) and our slope is 12\frac{1}{2}.

So let's use the format f(x)=m(xβˆ’x1)+y1f(x) = m(x-x_1) + y_1. Plugging in the numbers gets us:

f(x)=12(xβˆ’3)+34f(x) = \frac{1}{2}(x - 3) + \frac{3}{4}

This looks very close to our answer choices. By comparing it to the options, we can see that answer A is the right answer!

Conclusion

And there you have it! We've successfully found the linear function representing the arithmetic sequence. By understanding the relationship between arithmetic sequences and linear functions, we were able to use the given information to find the common difference (slope) and then determine the y-intercept. Always remember that each term in the arithmetic sequence represents a point on the line. I hope this was helpful. Keep practicing, and you'll become a pro at these problems in no time. Keep up the great work, and happy calculating!