Find The 55th Term Of The Arithmetic Sequence

So, picture this: I'm knee-deep in organizing my childhood photos. You know how it is, a massive box of blurry birthday parties and questionable fashion choices. I stumbled upon a stack of old report cards, and my eyes, naturally, went to the math grades. And then, I saw it. A particularly abysmal score on a test about… arithmetic sequences. My 12-year-old self was NOT thrilled. I remember thinking, "Why would I EVER need to know this stuff?" Well, fast forward a few decades, and here I am, about to tell you all about finding the 55th term of an arithmetic sequence. Funny how life works, right? It’s like the universe decided to give my younger self a subtle, long-overdue wink.

Now, before you click away thinking, "Oh no, math class is back!" – hear me out. This isn't about mind-numbingly complex equations or proving theorems. This is about a pattern. And who doesn't love a good pattern? Think of your favorite song, the one with that catchy chorus that repeats. Or that TV show where you can predict what the grumpy neighbor is going to say next. Arithmetic sequences are just like that, but with numbers!

Let's break it down. What exactly is an arithmetic sequence? Imagine you're counting sheep, but instead of counting by ones, you decide to count by twos. So you have 2, 4, 6, 8, and so on. Or maybe you're saving money, and you put aside $5 every week. So that's $5, $10, $15, $20… See? There's a consistent difference between each number. That consistent difference is the magic ingredient. It's what makes it an arithmetic sequence. Mathematicians, bless their organized hearts, call this the common difference, and they usually represent it with the letter 'd'.

Think of it like a train. Each car is a number in the sequence. And the engine that pulls them all along, adding that consistent distance between each car? That's your common difference. If you have the sequence 3, 7, 11, 15… what's the engine? Yep, it's 4. You add 4 to get from 3 to 7, you add 4 to get from 7 to 11, and so on. So, d = 4 in this case.

The Anatomy of an Arithmetic Sequence

Every arithmetic sequence has a starting point. This is the very first number in your line-up. We call this the first term, and it's usually represented by the letter 'a' with a little '1' subscript, like a₁. So, in our 3, 7, 11, 15 example, a₁ = 3.

Now, the cool thing is, once you know the first term and the common difference, you can pretty much figure out any number in that sequence. It's like having a cheat code for numbers. You don't need to painstakingly add the common difference over and over again. There’s a shortcut! And oh, how we all love a good shortcut, especially when it comes to math.

Let’s try to build a sequence from scratch. Let's say our first term, a₁, is 10. And let's pick a common difference, d, of 7. What would the sequence look like?

The first term is 10. Easy enough. a₁ = 10.

The second term, a₂, is the first term plus the common difference. So, 10 + 7 = 17. a₂ = 17.

The third term, a₃, is the second term plus the common difference. So, 17 + 7 = 24. a₃ = 24.

The fourth term, a₄, is the third term plus the common difference. So, 24 + 7 = 31. a₄ = 31.

And so on. We're getting 10, 17, 24, 31… You can see how if we wanted the 100th term, we’d be here all day just adding 7. That’s where the formula comes in, the superhero of arithmetic sequences!

The Magic Formula: Unlocking Any Term

Here it is, the big reveal! The formula to find the nth term of an arithmetic sequence (where 'n' is just the position of the term you're looking for) is:

a_n = a₁ + (n - 1)d

Let’s decode this. Remember:

• a_n is the term you want to find (the value of the number at that position).
• a₁ is the very first term in your sequence.
• n is the position of the term you are looking for (e.g., if you want the 55th term, then n = 55).
• d is the common difference – that consistent jump between numbers.

The (n - 1) part is super important. Why minus 1? Think about it. To get to the second term (n=2), you only add the common difference once to the first term. To get to the third term (n=3), you add the common difference twice. To get to the fourth term (n=4), you add it three times. So, to get to the nth term, you add the common difference (n - 1) times. It's like taking a certain number of steps; you take one less step to reach the end of a path than the total number of markers you pass.

Let's revisit our 10, 17, 24, 31… sequence. We know a₁ = 10 and d = 7. Let's say we want to find the 4th term (n = 4) using the formula. Just to check our work, right?

a₄ = a₁ + (4 - 1)d

a₄ = 10 + (3) * 7

a₄ = 10 + 21

a₄ = 31

Boom! It matches what we found by just adding. See? The formula works!

Finding the 55th Term: The Main Event!

Okay, drumroll please… let’s tackle our original mission: finding the 55th term of an arithmetic sequence. But wait, we need a sequence to work with! Let’s make one up. This is where we get to be the architects of our numerical destiny.

Let's say our arithmetic sequence starts with the number 5. So, a₁ = 5. And let's say the common difference is 3. So, d = 3.

Our sequence looks like: 5, 8, 11, 14, 17, 20… and it goes on, forever and ever (or at least, until we get tired of writing it down).

We want to find the 55th term. So, what’s our n? You guessed it: n = 55.

Now, plug these values into our trusty formula: a_n = a₁ + (n - 1)d

a₅₅ = 5 + (55 - 1) * 3

First, let's handle the subtraction inside the parentheses: 55 - 1 = 54.

So now we have: a₅₅ = 5 + (54) * 3

Next, we do the multiplication: 54 * 3. Let's do that math… 50 * 3 is 150, and 4 * 3 is 12. Add them up, and we get 162.

Our equation is now: a₅₅ = 5 + 162

And finally, the addition: 5 + 162 = 167.

So, the 55th term of the arithmetic sequence that starts with 5 and has a common difference of 3 is… 167!

Pretty neat, right? Imagine trying to add 3, fifty-four times, to the number 5. It's doable, but a bit tedious. The formula just gives us a direct route, saving us time and potential finger cramps.

When Things Get Tricky (But Not Too Tricky)

What if you're not given the first term directly? Or what if the common difference is negative? Does the formula still hold up? Absolutely!

Let's say you’re told: "The 10th term of an arithmetic sequence is 30, and the 15th term is 45. Find the 55th term."

Uh oh, missing a₁ and d. Don't panic! We can figure these out.

We know that the difference between the 15th term and the 10th term is made up of the common difference being added multiple times. How many times? Well, between the 10th and 15th term, there are 15 - 10 = 5 steps (or 5 common differences).

So, the difference in value (45 - 30 = 15) must equal 5 times the common difference (5d).

15 = 5d

Divide both sides by 5, and you get d = 3.

Awesome, we found the common difference! Now we need the first term. We can use our formula a_n = a₁ + (n - 1)d and plug in one of the terms we know. Let's use the 10th term (a₁₀ = 30).

30 = a₁ + (10 - 1) * 3

30 = a₁ + (9) * 3

30 = a₁ + 27

To find a₁, subtract 27 from both sides:

30 - 27 = a₁

3 = a₁

So, for this tricky scenario, our sequence starts with 3 and has a common difference of 3. Now, finding the 55th term is back to our familiar territory!

a₅₅ = a₁ + (n - 1)d

a₅₅ = 3 + (55 - 1) * 3

a₅₅ = 3 + (54) * 3

a₅₅ = 3 + 162

a₅₅ = 165

See? Even when it looks a little more complicated, the core principles and the formula are your best friends. They’re like the trusty Swiss Army knife of arithmetic.

Why Bother? (Besides My Report Card Shame)

Okay, I know what some of you might be thinking: "This is still just abstract numbers. Where does this ever come up in the real world?"

Well, besides the obvious uses in more advanced math, think about things that grow or diminish at a steady rate. If a plant grows 2 centimeters every day, that’s an arithmetic sequence. If you're paying off a loan, and you reduce the balance by a fixed amount each month, that’s an arithmetic sequence. If you’re training for a marathon and your coach tells you to run 500 meters more each week than the week before, hello arithmetic sequence!

It’s all about predictability and consistent change. And understanding these patterns helps us model and predict all sorts of phenomena. Plus, it’s a fantastic way to train your brain to think logically and solve problems step-by-step. It’s like a mental workout!

And hey, maybe if I'd paid more attention to my 12-year-old math teacher, I wouldn't have had that embarrassing report card moment. But then again, maybe I wouldn't have had the chance to rediscover this and share it with you all. Every experience, even a bad grade, has its own lesson, doesn't it?

So, next time you see a sequence of numbers with a consistent gap, don't shy away. Embrace it! Use the formula. Find that 55th term, or the 100th, or even the 1000th. You've got the power now. You’ve got the formula, and that, my friends, is a beautiful thing. It's the key to unlocking a hidden order in the world of numbers, one steady step at a time.