Series & Arithmetic Progression: Solved Math Problems

by ADMIN 54 views

Hey math enthusiasts! Ready to dive into some cool problems involving series and arithmetic progressions? Let's break down these concepts step by step. We'll find the number of terms, calculate sums, and explore the awesome world of sequences. Get your pencils and calculators ready; it's gonna be fun! I'll guide you through each problem, explaining the logic so you can tackle similar challenges with confidence. Let's get started!

1. Finding the Number of Terms and Sums of Series

Alright guys, first up, we have two series. The first one is an arithmetic series, where the difference between consecutive terms is constant. The second is a geometric series, where each term is multiplied by a constant ratio. Our mission? To find out how many terms are in each series and calculate their sums. Sounds like a plan?

(a) Arithmetic Series: 5 + 7 + 9 + 11 + ... + 41

Let's tackle this arithmetic series. We've got 5, 7, 9, 11, and it goes on until 41. The key here is recognizing the common difference, which is the constant value added to get from one term to the next. In this case, we're adding 2 each time. So, the common difference (d) is 2. Now, let's figure out how many terms are in this series. We'll use the formula for the nth term of an arithmetic series: an = a1 + (n - 1) * d. Where:

  • an = the last term (41)
  • a1 = the first term (5)
  • n = the number of terms (what we want to find!)
  • d = the common difference (2)

Let's plug in the numbers: 41 = 5 + (n - 1) * 2. Simplify this equation, and solve for 'n'. First, subtract 5 from both sides: 36 = (n - 1) * 2. Then, divide both sides by 2: 18 = n - 1. Finally, add 1 to both sides: n = 19. Boom! There are 19 terms in this series. Now that we know the number of terms, let's find the sum (S). We'll use the formula for the sum of an arithmetic series: S = n/2 * (a1 + an). Where:

  • n = the number of terms (19)
  • a1 = the first term (5)
  • an = the last term (41)

Plug in the numbers: S = 19/2 * (5 + 41). Calculate this, and you get: S = 19/2 * 46 = 19 * 23 = 437. So, the sum of this arithmetic series is 437. Awesome, right? We've successfully found both the number of terms and the sum of this series. You're doing great!

(b) Geometric Series: 3 + 6 + 12 + 24 + ... + 1536

Now, let's switch gears and look at a geometric series: 3, 6, 12, 24, and so on, ending at 1536. Here, instead of adding a constant value, we're multiplying by a constant ratio. To find this ratio (r), divide any term by its preceding term. For example, 6 / 3 = 2, 12 / 6 = 2, and 24 / 12 = 2. So, the common ratio (r) is 2. To find the number of terms (n), we use the formula for the nth term of a geometric series: an = a1 * r^(n-1). Where:

  • an = the last term (1536)
  • a1 = the first term (3)
  • r = the common ratio (2)
  • n = the number of terms (what we need to find!)

Let's plug in the values: 1536 = 3 * 2^(n-1). Divide both sides by 3: 512 = 2^(n-1). Since 512 is a power of 2, we can rewrite it as 2^9 = 2^(n-1). Therefore, 9 = n - 1. Add 1 to both sides: n = 10. So, there are 10 terms in this geometric series. To find the sum (S) of a geometric series, we use the formula: S = a1 * (r^n - 1) / (r - 1). Where:

  • a1 = the first term (3)
  • r = the common ratio (2)
  • n = the number of terms (10)

Plug in the values: S = 3 * (2^10 - 1) / (2 - 1). Calculate this: S = 3 * (1024 - 1) / 1 = 3 * 1023 = 3069. The sum of this geometric series is 3069. Wow, two series down! You're becoming a series superstar!

2. Arithmetic Progression Problems

Alright, let's get into the arithmetic progression problems. We'll be using what we know about arithmetic series to solve for missing terms and find sums. Get ready to flex those math muscles!

(a) Finding the Common Difference and Last Term

Here we have an arithmetic progression where the first term (a1) is 5, the 5th term is 30, and the last term is 205. Our first goal is to find the common difference (d). Remember, each term in an arithmetic progression is found by adding the common difference to the previous term. The formula for the nth term is: an = a1 + (n - 1) * d. We know the 5th term (a5) is 30, so let's use that information. We can write: 30 = 5 + (5 - 1) * d. Simplify: 30 = 5 + 4 * d. Subtract 5 from both sides: 25 = 4 * d. Divide both sides by 4: d = 6.25. So, the common difference is 6.25. Now we know how much we add to each term to get the next term. Our next task is to find out the number of terms in the sequence. To do this we can use the following approach. We know the first term (a1 = 5), the common difference (d = 6.25), and the last term (an = 205). We can use the formula again: an = a1 + (n - 1) * d. Plugging in the values we have: 205 = 5 + (n - 1) * 6.25. Subtract 5 from both sides: 200 = (n - 1) * 6.25. Then, divide both sides by 6.25: 32 = n - 1. Finally, add 1 to both sides: n = 33. This means that there are 33 terms in this arithmetic progression. We have all the information necessary to determine the sum of this sequence. Let's calculate the sum (S) using the formula: S = n/2 * (a1 + an). Where:

  • n = the number of terms (33)
  • a1 = the first term (5)
  • an = the last term (205)

Plug in the numbers: S = 33/2 * (5 + 205). Calculate this: S = 33/2 * 210 = 33 * 105 = 3465. The sum of this arithmetic progression is 3465. There you have it! By breaking down the problem into smaller steps, we successfully solved for the common difference, the number of terms, and the sum of the series. Keep practicing, and you'll master these concepts in no time.

(b) Finding the Sum of an Arithmetic Progression

In an arithmetic progression, the first term is 10, the last term is 50, and there are 8 terms in total. This one is pretty straightforward. We're given the first term (a1 = 10), the last term (an = 50), and the number of terms (n = 8). We need to find the sum (S). We can use the same formula we used before: S = n/2 * (a1 + an). Where:

  • n = the number of terms (8)
  • a1 = the first term (10)
  • an = the last term (50)

Plug in the values: S = 8/2 * (10 + 50). Calculate: S = 4 * 60 = 240. The sum of this arithmetic progression is 240. Easy peasy, right? Knowing the formulas and how to apply them makes these problems a breeze.

Conclusion

Great job, everyone! We've successfully worked through several series and arithmetic progression problems. We've learned how to find the number of terms, calculate sums, and understand the difference between arithmetic and geometric series. Remember, practice is key. The more you work through these types of problems, the more comfortable and confident you'll become. Keep up the amazing work, and happy calculating!