Arithmetic Progressions And Sum Of Multiples Step By Step Solutions

July 25, 2025 by Scholario Team 68 views

Unlocking Arithmetic Progressions and Multiples A Comprehensive Guide to Summation Problems

Hey there, math enthusiasts! Ever found yourself wrestling with arithmetic progressions (P.A.) or the sums of multiples within a range? You're not alone! These concepts can seem tricky, but with a little guidance, they become much easier to handle. In this article, we're diving deep into two specific problems and exploring the underlying principles. So, buckle up, and let's get started!

Cracking the Code of Arithmetic Progressions

Let's kick things off with a common type of problem that involves arithmetic progressions. Arithmetic progressions, for those who might need a refresher, are sequences of numbers where the difference between consecutive terms is constant. This constant difference is often referred to as the common difference. Understanding arithmetic progressions is crucial because they pop up in various mathematical contexts, from basic algebra to more advanced calculus. So, grasping the fundamentals here will definitely pay off in the long run. The beauty of arithmetic progressions lies in their predictability. Because the common difference remains the same throughout the sequence, we can easily determine any term in the progression or the sum of any number of terms. This predictability makes them a favorite in mathematical problems and real-world applications alike. Whether you're calculating simple interest, modeling linear growth, or even optimizing sequences in computer science, arithmetic progressions provide a powerful toolset.

Problem 1 Finding the Summation Terms

Our first challenge involves figuring out how many terms of a specific arithmetic progression we need to add up to reach a target sum. It’s like figuring out how many steps you need to take to reach a certain distance, where each step is a consistent length. Let's consider the arithmetic progression (-5, -1, 3, ...). The question we're tackling is: How many terms of this sequence must be summed, starting from the first term, to reach a sum of 1590? This problem isn't just about crunching numbers; it's about understanding the structure of arithmetic progressions and how their sums behave. We need to strategically apply the formulas that govern these sequences to unravel the mystery. Think of it as a detective trying to piece together clues to solve a case. Each term in the sequence, each difference, and the target sum all provide valuable information that, when combined, leads us to the solution. The fun part is putting these pieces together in the right way!

Breaking Down the Problem

To solve this, we'll need to dust off our knowledge of the arithmetic progression sum formula. This formula is a gem because it allows us to calculate the sum of a series of terms without having to add them up individually – especially handy when dealing with a large number of terms. The formula is typically expressed as: S_n = n/2 * [2a + (n - 1)d], where:

S_n is the sum of the first n terms,
n is the number of terms,
a is the first term of the progression, and
d is the common difference. This formula is the key to unlocking this problem, so let's make sure we understand each component. The sum S_n is what we're aiming for (1590 in this case). The number of terms n is what we're trying to find – the big unknown! The first term a is easily identifiable in our sequence (-5), and the common difference d can be calculated by subtracting any term from its successor. Once we have these pieces, it's just a matter of plugging them into the formula and solving for n. Think of it as baking a cake – each ingredient has its role, and when combined correctly, they create something delicious (or in this case, a mathematical solution!).

Putting the Formula to Work

In our specific arithmetic progression (-5, -1, 3, ...), the first term (a) is -5. To find the common difference (d), we subtract the first term from the second term: -1 - (-5) = 4. So, d is 4. We also know that S_n (the sum we want) is 1590. Now, we can substitute these values into the formula:

1590 = n/2 * [2(-5) + (n - 1)4]

This equation might look a bit intimidating at first, but don't worry! It's just an algebraic equation that we can solve using our trusty math skills. The next step is to simplify and rearrange the equation to make it more manageable. We'll distribute, combine like terms, and hopefully end up with a quadratic equation. Quadratic equations are a common sight in math problems, and there are several ways to solve them, including factoring, completing the square, or using the quadratic formula. The key here is to be systematic and careful with each step. Math is like building with LEGOs – each step builds upon the previous one, and a solid foundation is crucial for success.

Solving the Quadratic Equation

After simplifying the equation, we get:

1590 = n/2 * [-10 + 4n - 4]

1590 = n/2 * (4n - 14)

1590 = n * (2n - 7)

1590 = 2n² - 7n

Rearranging, we have a quadratic equation: 2n² - 7n - 1590 = 0.

Now, we can use the quadratic formula to solve for n: n = [-b ± √(b² - 4ac)] / (2a). In our equation, a = 2, b = -7, and c = -1590. Plugging these values into the formula gives us:

n = [7 ± √((-7)² - 4 * 2 * (-1590))] / (2 * 2)

n = [7 ± √(49 + 12720)] / 4

n = [7 ± √12769] / 4

n = [7 ± 113] / 4

This gives us two possible solutions for n: (n = 120 / 4 = 30) or (n = -106 / 4 = -26.5). Since the number of terms cannot be negative or a fraction, we discard the second solution. Therefore, we need to sum 30 terms of the arithmetic progression to reach a sum of 1590. Isn't it satisfying when the pieces finally fall into place? We started with a problem, broke it down into smaller, manageable steps, and used our mathematical tools to arrive at the answer. This process is what makes math so rewarding!

Summing Multiples Within a Range

Now, let's shift our focus to another intriguing type of problem: summing multiples within a given range. This involves identifying numbers that are divisible by a specific number (the multiple) and then adding them all up. These types of problems often appear in number theory and can be quite challenging if you don't have the right approach. However, fear not! We're here to break it down step by step. The key to success here is to recognize that the multiples of a number within a range also form an arithmetic progression. This realization allows us to leverage the same formulas and techniques we used earlier, making the problem much more approachable. Think of it as finding hidden patterns – once you see the pattern, the solution becomes clear. It's like discovering a secret code that unlocks the answer!

Problem 2 Sum of Multiples of 11

Our next challenge is to find the sum of all multiples of 11 that fall between 100 and 10000. This might seem like a daunting task at first – imagining adding up all those multiples individually! But, as we've hinted, there's a much more efficient way to tackle this. The beauty of mathematics is that it often provides us with shortcuts and elegant solutions to seemingly complex problems. It's all about finding the right tool for the job. In this case, our tool is the arithmetic progression formula, which, as we've seen, can be incredibly powerful when dealing with sequences of numbers that follow a consistent pattern. So, let's put on our thinking caps and get ready to apply this tool to solve our problem.

Identifying the Arithmetic Progression

The first step is to identify the smallest and largest multiples of 11 within our range (100 to 10000). The smallest multiple of 11 greater than 100 is 11 * 10 = 110. The largest multiple of 11 less than 10000 is 11 * 909 = 9999. This means our arithmetic progression consists of the numbers 110, 121, 132, ..., 9999. Spotting these boundaries is crucial because they define the start and end of our sequence. Without them, we'd be swimming in a sea of numbers without a clear direction. Think of it as setting the boundaries of a game – you need to know where the playing field begins and ends before you can strategize and play effectively.

Now, we need to determine how many terms are in this progression. We can use the formula for the nth term of an arithmetic progression: a_n = a + (n - 1)d, where:

a_n is the nth term (9999 in our case),
a is the first term (110),
n is the number of terms (what we want to find), and
d is the common difference (11).

Plugging in the values, we get:

9999 = 110 + (n - 1)11

Calculating the Number of Terms

Let's solve for n:

9999 - 110 = (n - 1)11

9889 = (n - 1)11

9889 / 11 = n - 1

899 = n - 1

n = 900

So, there are 900 multiples of 11 between 100 and 10000. This is a significant number, highlighting the importance of using a formula rather than manually adding up all the terms. Imagine trying to add 900 numbers together – it would be a tedious and error-prone process! This is where the power of mathematical tools truly shines – they allow us to tackle complex problems with efficiency and accuracy.

Finding the Sum

Now that we know the first term (110), the last term (9999), and the number of terms (900), we can use the arithmetic series sum formula again. This time, we can use a slightly different version of the formula that's particularly handy when we know the last term: S_n = n/2 * (a + a_n). Plugging in the values:

S_900 = 900 / 2 * (110 + 9999)

S_900 = 450 * 10109

S_900 = 4549050

Therefore, the sum of all multiples of 11 between 100 and 10000 is 4,549,050. Wow, that's a big number! But thanks to our methodical approach and the power of arithmetic progression formulas, we were able to calculate it without breaking a sweat. This problem beautifully illustrates how seemingly complex calculations can be simplified by recognizing underlying patterns and applying the appropriate mathematical tools.

Problem 3 Sum of Positive Multiples of 7

Finally, let's tackle a slightly broader problem: finding the sum of all positive multiples of 7 with two, three, or four digits. This problem adds an extra layer of complexity because we need to consider multiple ranges. But don't worry, we'll break it down step by step, just like we did before. The key here is to divide and conquer. We'll treat each range (two-digit, three-digit, and four-digit) separately, calculate the sum of multiples within each range, and then add those sums together. This approach makes the problem much more manageable and less overwhelming.

Defining the Ranges

First, let's define the ranges:

Two-digit multiples of 7: between 10 and 99
Three-digit multiples of 7: between 100 and 999
Four-digit multiples of 7: between 1000 and 9999

Now, we'll apply the same strategy we used in the previous problem: identify the first and last multiple of 7 within each range, determine the number of terms, and then use the arithmetic series sum formula to calculate the sum. This consistent approach is crucial for solving these types of problems efficiently. By sticking to a well-defined strategy, we can avoid confusion and ensure that we cover all the necessary steps.

Two-Digit Multiples of 7

The smallest two-digit multiple of 7 is 7 * 2 = 14, and the largest is 7 * 14 = 98. To find the number of terms, we can use the same logic as before. The multiples of 7 in this range are 14, 21, 28, ..., 98. We can think of this as an arithmetic progression with a first term of 14, a common difference of 7, and a last term of 98. Using the formula a_n = a + (n - 1)d:

98 = 14 + (n - 1)7

Solving for n:

98 - 14 = (n - 1)7

84 = (n - 1)7

84 / 7 = n - 1

12 = n - 1

n = 13

So, there are 13 two-digit multiples of 7. Now, we can find the sum using the formula S_n = n/2 * (a + a_n):

S_13 = 13 / 2 * (14 + 98)

S_13 = 6.5 * 112

S_13 = 728

Three-Digit Multiples of 7

The smallest three-digit multiple of 7 is 7 * 15 = 105, and the largest is 7 * 142 = 994. Using the same approach, we can find the number of terms:

994 = 105 + (n - 1)7

Solving for n:

994 - 105 = (n - 1)7

889 = (n - 1)7

889 / 7 = n - 1

127 = n - 1

n = 128

So, there are 128 three-digit multiples of 7. The sum is:

S_128 = 128 / 2 * (105 + 994)

S_128 = 64 * 1099

S_128 = 70336

Four-Digit Multiples of 7

The smallest four-digit multiple of 7 is 7 * 143 = 1001, and the largest is 7 * 1428 = 9996. The number of terms is:

9996 = 1001 + (n - 1)7

Solving for n:

9996 - 1001 = (n - 1)7

8995 = (n - 1)7

8995 / 7 = n - 1

1285 = n - 1

n = 1286

The sum is:

S_1286 = 1286 / 2 * (1001 + 9996)

S_1286 = 643 * 10997

S_1286 = 7071071

The Final Sum

Finally, we add the sums from each range to get the total sum:

Total Sum = 728 + 70336 + 7071071 = 7142135

Therefore, the sum of all positive multiples of 7 with two, three, or four digits is 7,142,135. This problem demonstrates the power of breaking down complex problems into smaller, more manageable parts. By tackling each range separately and then combining the results, we were able to arrive at the solution without getting bogged down in the details.

Final Thoughts

And there you have it! We've successfully navigated through problems involving arithmetic progressions and sums of multiples. These types of problems might seem intimidating at first, but with the right tools and a systematic approach, they become much more approachable. Remember, the key is to understand the underlying principles, identify patterns, and apply the appropriate formulas. Math is like a puzzle – each problem is a new challenge, and the satisfaction of solving it is incredibly rewarding. So, keep practicing, keep exploring, and keep unlocking the amazing world of mathematics!

I hope this comprehensive guide has helped you better understand these concepts and feel more confident in tackling similar problems. Happy calculating, guys!