Geometric Series Sum Find The First Five Terms

by ADMIN 47 views

Hey guys! Today, we're diving deep into the fascinating world of geometric series. We're going to tackle a classic problem: finding the sum of the first five terms of a geometric series where the first term ($a_1$) is 6 and the common ratio ($r$) is 1/3. Buckle up, because we're not just going to find the answer โ€“ we're going to understand the why behind it all. And don't worry, we'll express our final answer as an improper fraction in lowest terms, nice and tidy.

Understanding Geometric Series: The Building Blocks

Before we jump into the specific problem, let's make sure we're all on the same page about what a geometric series actually is. Think of it as a sequence of numbers where each term is found by multiplying the previous term by a constant value. This constant value is our superstar, the common ratio, often denoted by r. So, if you have a first term ($a_1$), the second term is $a_1 * r$, the third is $a_1 * r^2$, and so on. You're essentially building a series by repeatedly multiplying by the same ratio.

Now, let's break this down even further. Imagine you start with the number 2 and your common ratio is 3. The first few terms of your geometric series would be: 2, 6, 18, 54, and so on. See how each number is simply the previous one multiplied by 3? That's the magic of a geometric series in action! You can easily find any term in the sequence if you know the first term and the common ratio.

But what if we want to find the sum of a certain number of terms in the series? That's where things get really interesting! Adding up the terms one by one can be tedious, especially if we're dealing with a large number of terms. Fortunately, there's a handy formula that makes our lives much easier. This formula is the key to unlocking the sum of any finite geometric series, and we'll use it to solve our problem today. The formula allows us to bypass the tedious addition and go straight to the answer. It's like having a secret code that unlocks the mystery of the series sum!

The Formula That Unlocks the Sum: A Deep Dive

Here comes the star of the show: the formula for the sum of the first n terms of a geometric series. It looks a little intimidating at first, but trust me, it's your best friend in these situations. The formula is:

Sn=a1โˆ—(1โˆ’rn)/(1โˆ’r)S_n = a_1 * (1 - r^n) / (1 - r)

Where:

  • S_n$ is the sum of the first *n* terms

  • a_1$ is the first term of the series

  • r is the common ratio
  • n is the number of terms we want to sum

Let's break down this formula piece by piece. The term $(1 - r^n)$ represents the difference between 1 and the common ratio raised to the power of the number of terms. This part of the formula captures the effect of the common ratio on the overall sum. The denominator, $(1 - r)$, normalizes the sum based on how much the terms are changing due to the common ratio. It's like a balancing factor that ensures our sum is accurate.

So, why does this formula work? Well, it's derived using a clever algebraic trick. Imagine writing out the sum of the first n terms and then multiplying the entire sum by r. When you subtract these two equations, a lot of terms cancel out, leaving you with a simplified expression that leads directly to the formula. The beauty of this formula lies in its efficiency. Instead of adding up numerous terms individually, we can simply plug in the values of $a_1$, r, and n and get the answer in one fell swoop. It's a powerful tool in our mathematical arsenal!

Applying the Formula to Our Problem: Let's Get Solving!

Now, let's get back to our original problem. We need to find the sum of the first five terms of a geometric series with $a_1 = 6$ and $r = 1/3$. That means n = 5. We have all the ingredients we need to bake up our answer using the formula. All we need to do is substitute the given values into the formula and simplify. It's like following a recipe โ€“ just plug in the numbers and let the math do its magic!

Let's substitute these values into our formula:

S5=6โˆ—(1โˆ’(1/3)5)/(1โˆ’1/3)S_5 = 6 * (1 - (1/3)^5) / (1 - 1/3)

Okay, it looks a bit intimidating with all those fractions and exponents, but let's take it step by step. First, we need to calculate $(1/3)^5$. This means 1/3 multiplied by itself five times: $(1/3) * (1/3) * (1/3) * (1/3) * (1/3) = 1/243$. Remember, when raising a fraction to a power, we raise both the numerator and the denominator to that power.

Now, let's plug that back into our equation:

S5=6โˆ—(1โˆ’1/243)/(1โˆ’1/3)S_5 = 6 * (1 - 1/243) / (1 - 1/3)

Next, we need to deal with the expressions inside the parentheses. Let's start with $(1 - 1/243)$. To subtract these, we need a common denominator, which is 243. So, we rewrite 1 as 243/243. Now we have:

(243/243โˆ’1/243)=242/243(243/243 - 1/243) = 242/243

Moving on to the denominator, we have $(1 - 1/3)$. Again, we need a common denominator, which is 3. So, we rewrite 1 as 3/3. Now we have:

(3/3โˆ’1/3)=2/3(3/3 - 1/3) = 2/3

Our equation now looks like this:

S5=6โˆ—(242/243)/(2/3)S_5 = 6 * (242/243) / (2/3)

We're getting closer! Now we have a fraction divided by another fraction. Remember the rule for dividing fractions? We invert the second fraction and multiply. So, dividing by 2/3 is the same as multiplying by 3/2. Let's rewrite our equation:

S5=6โˆ—(242/243)โˆ—(3/2)S_5 = 6 * (242/243) * (3/2)

Simplifying to the Final Answer: The Grand Finale

Before we start multiplying these fractions, let's look for opportunities to simplify. Simplifying fractions before multiplying makes the calculations much easier. We can see that 6 and 2 have a common factor of 2. Let's divide both by 2: 6 becomes 3 and 2 becomes 1. We can also see that 3 and 243 have a common factor of 3. Let's divide both by 3: 3 becomes 1 and 243 becomes 81.

Our equation now looks like this:

S5=3โˆ—(242/81)โˆ—(1/1)S_5 = 3 * (242/81) * (1/1)

Now we can multiply the numerators and the denominators:

S5=(3โˆ—242โˆ—1)/(1โˆ—81โˆ—1)=726/81S_5 = (3 * 242 * 1) / (1 * 81 * 1) = 726 / 81

We've got our answer as an improper fraction, but we're not quite done yet. We need to express it in lowest terms. This means finding the greatest common divisor (GCD) of 726 and 81 and dividing both the numerator and the denominator by it.

Both 726 and 81 are divisible by 3. Dividing 726 by 3 gives us 242, and dividing 81 by 3 gives us 27. So, our fraction becomes:

S5=242/27S_5 = 242 / 27

Can we simplify further? Let's check. The factors of 27 are 1, 3, 9, and 27. The factors of 242 are 1, 2, 11, 22, 121, and 242. The only common factor is 1, so our fraction is already in its simplest form.

Therefore, the sum of the first five terms of the geometric series is 242/27. We did it! We successfully applied the formula, simplified our fractions, and arrived at the final answer, expressed as an improper fraction in lowest terms. Give yourselves a pat on the back โ€“ you've conquered the geometric series! This problem showcases how a seemingly complex concept can be broken down into manageable steps with the right tools and understanding.