Finding The Nth Term Of An Arithmetic Progression

Hey guys! Let's dive into a super common and important type of math problem – finding the nth term of an arithmetic progression (AP). You'll often see these in exams, especially the ENEM (Exame Nacional do Ensino Médio) in Brazil, so it’s crucial to get the hang of them. Today, we're tackling a problem where we know the sum of the first 'n' terms, the first term itself, and the common difference. Our mission? To figure out the value of that nth term. Let's break it down step-by-step so it’s crystal clear.

Understanding Arithmetic Progression (AP)

First things first, let's make sure we're all on the same page about what an arithmetic progression actually is. An arithmetic progression, or AP, is basically a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is what we call the common difference, often denoted by 'r'.

Think of it like this: You start with a number (the first term, ${ a_1 }$), and then you keep adding the same number ('r') to get the next term. So, if your first term is 2 and your common difference is 3, your sequence would look like this: 2, 5, 8, 11, and so on. Each term is simply the previous term plus 3.

In mathematical terms, we can write the nth term of an AP using a simple formula:

${ a_n = a_1 + (n - 1) \times r }$

Where:

• ${ a_n }$ is the nth term (the term we want to find).
• ${ a_1 }$ is the first term of the sequence.
• ${ n }$ is the number of terms.
• ${ r }$ is the common difference.

This formula is super handy because it lets us jump straight to any term in the sequence without having to list out all the terms before it. For instance, if we wanted to find the 100th term of the sequence 2, 5, 8, 11..., we could use this formula directly instead of adding 3 a hundred times!

Another key concept we need to understand is the sum of the first 'n' terms of an AP. There's a formula for that too, which can save us a lot of time when we're dealing with longer sequences. The sum, often denoted by ${ S_n }$, is given by:

${ S_n = \frac{n}{2} [2a_1 + (n - 1) \times r] }$

Where:

• ${ S_n }$ is the sum of the first 'n' terms.
• ${ n }$ is the number of terms.
• ${ a_1 }$ is the first term.
• ${ r }$ is the common difference.

This formula might look a bit intimidating at first, but it's really just a way of adding up all the terms in the AP in a more efficient way. It's derived from the idea of pairing the first and last terms, the second and second-to-last terms, and so on – each pair adds up to the same value, which makes the calculation much simpler. Understanding these formulas and how they work is the first big step in tackling AP problems. So, make sure you've got these down before we move on to solving our specific problem!

Problem Breakdown: Sum of the First 'n' Terms

Okay, let's break down the problem we're facing. We know a few key pieces of information about our arithmetic progression:

• The sum of the first 'n' terms, ${ S_n }$, is 205.
• The first term, ${ a_1 }$, is 7.
• The common difference, ${ r }$, is 3.

Our ultimate goal is to find the nth term, ${ a_n }$. But here's the catch: we don't know what 'n' is! That's the number of terms in the AP, and we need to figure it out before we can find ${ a_n }$. This is a classic problem-solving situation where we have to use the information we have to find the missing piece of the puzzle.

The fact that we know the sum of the first 'n' terms is a big clue. It points us directly to the formula for the sum of an AP, which we discussed earlier:

${ S_n = \frac{n}{2} [2a_1 + (n - 1) \times r] }$

We can plug in the values we know – ${ S_n = 205 }$, ${ a_1 = 7 }$, and ${ r = 3 }$ – into this formula. This will give us an equation with 'n' as the only unknown. Once we solve for 'n', we'll know exactly how many terms are in the sequence, which is a critical step towards finding the nth term itself.

So, the plan is clear: we're going to use the sum formula to create an equation, solve that equation to find 'n', and then use 'n' to find ${ a_n }$. This is a common strategy in math problems – using one formula to find a missing value that we can then use in another formula. It's like a chain reaction, where each step unlocks the next. Now, let's get into the nitty-gritty of plugging in the values and solving for 'n'. This is where the algebra comes in, but don't worry, we'll take it slow and steady!

Plugging in the Values

Let's get our hands dirty with some algebra! We're going to take the sum formula and replace the symbols with the numbers we know. Remember, we have:

• ${ S_n = 205 }$
• ${ a_1 = 7 }$
• ${ r = 3 }$

Plugging these into the sum formula, we get:

${ 205 = \frac{n}{2} [2 \times 7 + (n - 1) \times 3] }$

See how we've replaced ${ S_n }$, ${ a_1 }$, and ${ r }$ with their respective values? Now we have an equation that only involves 'n'. This is a big step forward because we can now focus on isolating 'n' and finding its value. But before we jump into solving, let's simplify the equation a bit. This will make our lives easier in the long run.

First, let's deal with the stuff inside the brackets. We can multiply 2 by 7 to get 14, and we can distribute the 3 across the ${ (n - 1) }$ term. That means we multiply 3 by 'n' and 3 by -1. This gives us:

${ 205 = \frac{n}{2} [14 + 3n - 3] }$

Now we can simplify further by combining the constants inside the brackets. We have 14 and -3, which add up to 11. So our equation becomes:

${ 205 = \frac{n}{2} [11 + 3n] }$

We're getting there! The equation is looking cleaner and more manageable. The next step is to get rid of that fraction. We have ${ \frac{n}{2} }$ multiplying the brackets, so to undo that division by 2, we can multiply both sides of the equation by 2. This is a crucial step in solving for 'n' because it clears the fraction and makes the equation much easier to work with. So, let's multiply both sides by 2 and see what we get.

Simplifying the Equation

Okay, let's continue simplifying our equation. We're at:

${ 205 = \frac{n}{2} [11 + 3n] }$

As we discussed, we want to get rid of the fraction, so we'll multiply both sides of the equation by 2. This is a valid algebraic operation because what we do to one side, we must do to the other to keep the equation balanced.

Multiplying both sides by 2 gives us:

${ 2 \times 205 = n [11 + 3n] }$

On the left side, ${ 2 \times 205 }$ is simply 410. On the right side, the 2 in the denominator has been cancelled out, leaving us with 'n' multiplied by the expression in the brackets. So our equation now looks like this:

${ 410 = n [11 + 3n] }$

Now, let's distribute the 'n' on the right side. This means we multiply 'n' by both 11 and 3n. Remember that ${ n \times 3n }$ is ${ 3n^2 }$. So, after distributing, we get:

${ 410 = 11n + 3n^2 }$

This is starting to look like a quadratic equation! A quadratic equation is an equation of the form ${ ax^2 + bx + c = 0 }$, where 'a', 'b', and 'c' are constants. Our equation has a term with ${ n^2 }$, a term with 'n', and a constant term, so it fits the bill. To solve a quadratic equation, we usually want to get it into the standard form, which means having all the terms on one side and zero on the other. Let's rearrange our equation to do just that.

We can subtract 410 from both sides to get everything on the right side and zero on the left. This gives us:

${ 0 = 3n^2 + 11n - 410 }$

Great! We now have our equation in the standard quadratic form. This is a big step because we have several methods we can use to solve for 'n'. The most common methods are factoring, using the quadratic formula, or completing the square. In this case, the quadratic formula is likely the most straightforward approach. So, let's get ready to apply the quadratic formula and find the values of 'n' that satisfy this equation.

Solving the Quadratic Equation

Alright, we've arrived at a quadratic equation:

${ 0 = 3n^2 + 11n - 410 }$

To solve this, we're going to use the quadratic formula. If you remember, the quadratic formula is a general formula for finding the solutions (also called roots) of any quadratic equation in the form ${ ax^2 + bx + c = 0 }$. The formula is:

${ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }$

Where:

• 'a' is the coefficient of the ${ n^2 }$ term.
• 'b' is the coefficient of the 'n' term.
• 'c' is the constant term.

In our equation, we can identify the coefficients as follows:

• ${ a = 3 }$
• ${ b = 11 }$
• ${ c = -410 }$

Now, we just need to plug these values into the quadratic formula and simplify. It might look a bit daunting with all the symbols, but it's really just a matter of careful substitution and arithmetic. So, let's do it!

Substituting the values into the formula, we get:

${ n = \frac{-11 \pm \sqrt{11^2 - 4 \times 3 \times (-410)}}{2 \times 3} }$

Now we need to simplify the expression. Let's start with the part under the square root. We have ${ 11^2 }$, which is 121. Then we have ${ -4 \times 3 \times (-410) }$, which is ${ 4920 }$. So the expression under the square root becomes ${ 121 + 4920 }$, which is 5041.

Our equation now looks like this:

${ n = \frac{-11 \pm \sqrt{5041}}{6} }$

The square root of 5041 is 71. So we have:

${ n = \frac{-11 \pm 71}{6} }$

Now we have two possible solutions for 'n', one where we add 71 and one where we subtract 71. Let's calculate both of them.

• Case 1: Addition

${ n = \frac{-11 + 71}{6} = \frac{60}{6} = 10 }$

• Case 2: Subtraction

${ n = \frac{-11 - 71}{6} = \frac{-82}{6} \approx -13.67 }$

So we have two possible values for 'n': 10 and approximately -13.67. But remember, 'n' represents the number of terms in the arithmetic progression. Can we have a negative or fractional number of terms? Nope! The number of terms must be a positive integer. Therefore, the only valid solution is ${ n = 10 }$. We've finally found 'n'! This means there are 10 terms in our arithmetic progression. Now that we know 'n', we're just one step away from finding the nth term, ${ a_n }$. We'll use the formula for the nth term, which we discussed earlier, and plug in the values we know. Let's do that next.

Determining the nth Term

Fantastic! We've successfully found that ${ n = 10 }$. This means we're looking for the 10th term of the arithmetic progression. Now, let's recall the formula for the nth term of an AP:

${ a_n = a_1 + (n - 1) \times r }$

We know:

• ${ a_1 = 7 }$ (the first term)
• ${ n = 10 }$ (the number of terms)
• ${ r = 3 }$ (the common difference)

Our goal is to find ${ a_{10} }$, which is the 10th term. So, let's plug the values we know into the formula:

${ a_{10} = 7 + (10 - 1) \times 3 }$

Now, we just need to simplify this expression. First, let's deal with the parentheses. ${ 10 - 1 }$ is 9. So our equation becomes:

${ a_{10} = 7 + 9 \times 3 }$

Next, we perform the multiplication. ${ 9 \times 3 }$ is 27. So we have:

${ a_{10} = 7 + 27 }$

Finally, we add 7 and 27, which gives us 34.

${ a_{10} = 34 }$

So, the 10th term of the arithmetic progression is 34. We did it! We've successfully found ${ a_n }$, which in this case is ${ a_{10} }$. This means that the tenth number in the sequence is 34. This was a multi-step problem, but we broke it down piece by piece, and now we have our answer. Let's take a moment to recap the steps we took to solve this problem. This will help solidify our understanding and make sure we can tackle similar problems in the future.

Wrapping Up: Putting It All Together

Okay, guys, let's take a step back and recap what we've done. We started with a problem where we knew the sum of the first 'n' terms of an arithmetic progression, the first term, and the common difference. Our mission was to find the nth term of this AP.

Here’s a quick rundown of the steps we took:

1. Understanding the Problem: We identified what we knew (${ S_n }$, ${ a_1 }$, ${ r }$) and what we needed to find (${ a_n }$).

2. Using the Sum Formula: We plugged the given values into the formula for the sum of an AP:

   ${ S_n = \frac{n}{2} [2a_1 + (n - 1) \times r] }$

3. Simplifying and Solving for 'n':

   • We simplified the equation and ended up with a quadratic equation.
   • We used the quadratic formula to solve for 'n'.
   • We discarded the negative solution because the number of terms must be a positive integer.
   • We found that ${ n = 10 }$.

4. Finding the nth Term:

   • We used the formula for the nth term of an AP:

     ${ a_n = a_1 + (n - 1) \times r }$

   • We plugged in the values for ${ a_1 }$, ${ n }$, and ${ r }$.

   • We calculated ${ a_{10} = 34 }$.

So, the 10th term of the arithmetic progression is 34. Phew! That was quite a journey, but we made it through. This problem is a great example of how we can combine different formulas and concepts to solve a more complex problem. It also highlights the importance of breaking down a problem into smaller, more manageable steps. When we tackle problems step-by-step, they become much less intimidating. Plus, practice makes perfect! The more you work through these types of problems, the more comfortable you'll become with the formulas and the problem-solving process. So, keep practicing, and you'll be an AP master in no time!

Remember, guys, math isn't about memorizing formulas; it's about understanding the concepts and how they connect. If you can understand the 'why' behind the formulas, you'll be able to apply them in all sorts of situations. Keep up the great work, and I'll catch you in the next math adventure!