Consecutive Numbers: Sum And Square Problem Solved!

by Sebastian Müller 52 views

Hey guys! Today, we're diving into a fun math problem that involves finding three consecutive numbers in an arithmetic progression (AP). It's like a little puzzle, and we're going to crack it together. The problem states that these three numbers add up to 36, and the square of the largest number is 196. Sounds interesting, right? Let's break it down step by step.

Understanding Arithmetic Progression

First off, let's make sure we're all on the same page about what an arithmetic progression is. An arithmetic progression is simply a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the 'common difference'. Think of it like a staircase where each step is the same height. For example, 2, 4, 6, 8... is an AP with a common difference of 2. Easy peasy!

In our case, we have three consecutive numbers in AP. So, if we call the first number 'a', the next number would be 'a + d', and the third would be 'a + 2d', where 'd' is our common difference. Remember, 'consecutive' here means they follow each other in order, like 1, 2, 3 or 10, 11, 12. The beauty of using this representation is that it neatly captures the relationship between the numbers in the progression.

Now, with the basics covered, we're ready to translate the word problem into mathematical equations. This is where the fun really begins! We have two key pieces of information: the sum of the numbers and the square of the largest number. Let's see how we can turn these into equations.

Setting Up the Equations

The first clue is that the sum of our three numbers is 36. So, we can write our first equation as:

a + (a + d) + (a + 2d) = 36

This equation represents the sum of the three consecutive numbers in our arithmetic progression. We've simply added the expressions we defined earlier for each number and set the total equal to 36. Now, let's simplify this equation a bit. Combining the 'a' terms, we get 3a. Combining the 'd' terms, we get 3d. So our simplified equation looks like this:

3a + 3d = 36

Cool, right? We've got our first equation. Now, let's tackle the second piece of information. The problem tells us that the square of the largest number is 196. Remember, our largest number is 'a + 2d'. So, we can write our second equation as:

(a + 2d)² = 196

This equation represents the square of the largest number in our arithmetic progression. We've taken the expression for the largest number and squared it, setting the result equal to 196. This equation looks a bit more complex than our first one, but don't worry, we'll handle it. With these two equations in hand, we're well on our way to solving the puzzle! We've successfully translated the word problem into the language of mathematics.

Solving the System of Equations

Alright, we've got our two equations:

  1. 3a + 3d = 36
  2. (a + 2d)² = 196

Now comes the exciting part – solving them! We have a system of equations, one linear and one quadratic. There are a few ways we can approach this. One common method is to solve the linear equation for one variable and then substitute that expression into the quadratic equation. This will give us a single equation with one variable, which we can then solve.

Let's start by simplifying our first equation, 3a + 3d = 36. We can divide both sides of the equation by 3 to make it simpler:

a + d = 12

Now, let's solve this equation for 'a'. We can subtract 'd' from both sides:

a = 12 - d

Great! We have an expression for 'a' in terms of 'd'. Now we can substitute this expression into our second equation, (a + 2d)² = 196. Replacing 'a' with '12 - d', we get:

((12 - d) + 2d)² = 196

Let's simplify the expression inside the parentheses:

(12 + d)² = 196

Now we have a quadratic equation in terms of 'd'. To solve it, we can take the square root of both sides:

√(12 + d)² = ±√196

This gives us two possible equations:

  1. 12 + d = 14
  2. 12 + d = -14

Let's solve each of these equations for 'd'. For the first equation, we subtract 12 from both sides:

d = 14 - 12

d = 2

For the second equation, we also subtract 12 from both sides:

d = -14 - 12

d = -26

So, we have two possible values for 'd': 2 and -26. Now we need to find the corresponding values for 'a'. We can use our expression a = 12 - d to do this.

Finding 'a' and the Numbers

Let's start with d = 2. Plugging this into a = 12 - d, we get:

a = 12 - 2

a = 10

So, if d = 2, then a = 10. This gives us the three numbers in the arithmetic progression:

  • a = 10
  • a + d = 10 + 2 = 12
  • a + 2d = 10 + 2(2) = 14

So, one set of numbers is 10, 12, and 14. Let's check if these numbers satisfy our original conditions. Their sum is 10 + 12 + 14 = 36, which is correct. The square of the largest number, 14², is 196, which is also correct. So, this set of numbers works!

Now, let's consider the other value for 'd', which is d = -26. Plugging this into a = 12 - d, we get:

a = 12 - (-26)

a = 12 + 26

a = 38

So, if d = -26, then a = 38. This gives us another set of three numbers in the arithmetic progression:

  • a = 38
  • a + d = 38 + (-26) = 12
  • a + 2d = 38 + 2(-26) = -14

So, our second set of numbers is 38, 12, and -14. Let's check these numbers as well. Their sum is 38 + 12 + (-14) = 36, which is correct. The square of the largest number in this case is 38, and 38² is not 196, so this set does not satisfy the condition that the square of the largest number is 196. Therefore, we can discard this solution.

The Solution

Awesome! We've solved the puzzle. The three consecutive numbers in arithmetic progression that sum up to 36 and have the square of the largest number equal to 196 are 10, 12, and 14.

Key Takeaways

This problem was a great example of how we can use algebra to solve real-world problems. The key steps we took were:

  1. Representing the numbers: We represented the three consecutive numbers in AP using variables 'a' and 'd', making it easier to form equations.
  2. Setting up equations: We translated the given information into mathematical equations, which is a crucial step in problem-solving.
  3. Solving the system: We used substitution to solve the system of equations and find the values of 'a' and 'd'.
  4. Checking the solution: We made sure to check our solution against the original conditions to ensure it was correct.

Remember, practice makes perfect! The more you solve these kinds of problems, the more comfortable you'll become with the process. Keep up the great work, guys! If you feel like doing other math problems or need help with other math stuff, let me know.