Solving Perimeter Of Rectangular Square And Analyzing Number Sequences

by Scholario Team 71 views

Hey there, math enthusiasts! Today, we're diving into a couple of intriguing problems that involve the world of geometry and number patterns. Get ready to put on your thinking caps as we unravel the mysteries behind a rectangular square's dimensions and a captivating numerical sequence.

Cracking the Rectangular Square's Perimeter

Let's kick things off with our rectangular square conundrum. We're told that the perimeter of this square is 180 meters, and here's the twist: the width is 10 meters shorter than the length. Our mission, should we choose to accept it, is to calculate the length and width of this square. Sounds like a fun challenge, right?

To tackle this problem head-on, we'll need to dust off our algebraic skills. Let's represent the length of the square with the variable "L" and the width with "W." Now, we can translate the given information into mathematical equations. We know that the perimeter of a rectangle is calculated by adding up all its sides, which can be expressed as 2L + 2W. So, our first equation is:

2L + 2W = 180

Next, we're told that the width is 10 meters less than the length. This can be written as:

W = L - 10

Now, we have a system of two equations with two unknowns, which is perfect for solving! We can use the substitution method to eliminate one of the variables. Let's substitute the second equation (W = L - 10) into the first equation:

2L + 2(L - 10) = 180

Now, it's time to simplify and solve for L:

2L + 2L - 20 = 180

4L - 20 = 180

4L = 200

L = 50

Great! We've found that the length (L) of the square is 50 meters. Now, we can use this value to find the width (W) using the equation W = L - 10:

W = 50 - 10

W = 40

So, there you have it! The length of the rectangular square is 50 meters, and the width is 40 meters. We've successfully cracked this geometric puzzle using the power of algebra. Give yourselves a pat on the back, guys!

Decoding the Numerical Sequence

Now, let's shift our focus to the second part of our challenge: the numerical sequence. We're presented with the following sequence: 2, 6, 12, 20, 30, ... The question is: what's the pattern here? What kind of sequence is this, and how can we predict the next numbers in line?

At first glance, this sequence might seem a bit mysterious. But fear not, my fellow math detectives! We can unravel its secrets by carefully examining the differences between consecutive terms. Let's calculate these differences:

  • 6 - 2 = 4
  • 12 - 6 = 6
  • 20 - 12 = 8
  • 30 - 20 = 10

Do you notice a pattern emerging? The differences between consecutive terms are increasing by 2 each time (4, 6, 8, 10). This suggests that the sequence is not a simple arithmetic sequence (where the difference between terms is constant), but rather a sequence with a more complex pattern.

To further investigate, let's look at the differences between these differences (the second-order differences):

  • 6 - 4 = 2
  • 8 - 6 = 2
  • 10 - 8 = 2

Aha! The second-order differences are constant and equal to 2. This is a key clue! It tells us that the sequence is a quadratic sequence, meaning that the general term of the sequence can be expressed as a quadratic equation of the form:

An = an^2 + bn + c

where An represents the nth term in the sequence, and a, b, and c are constants that we need to determine.

To find these constants, we can use the first few terms of the sequence and set up a system of equations. Let's use the first three terms (2, 6, 12):

  • For n = 1, A1 = 2: a(1)^2 + b(1) + c = 2 => a + b + c = 2
  • For n = 2, A2 = 6: a(2)^2 + b(2) + c = 6 => 4a + 2b + c = 6
  • For n = 3, A3 = 12: a(3)^2 + b(3) + c = 12 => 9a + 3b + c = 12

Now we have a system of three equations with three unknowns. We can solve this system using various methods, such as substitution or elimination. Let's use elimination. Subtracting the first equation from the second and the second equation from the third, we get:

  • (4a + 2b + c) - (a + b + c) = 6 - 2 => 3a + b = 4
  • (9a + 3b + c) - (4a + 2b + c) = 12 - 6 => 5a + b = 6

Now, subtract the first of these new equations from the second:

  • (5a + b) - (3a + b) = 6 - 4 => 2a = 2 => a = 1

We've found that a = 1! Now, we can substitute this value back into the equation 3a + b = 4 to find b:

3(1) + b = 4

3 + b = 4

b = 1

So, b = 1 as well. Finally, we can substitute the values of a and b into the equation a + b + c = 2 to find c:

1 + 1 + c = 2

2 + c = 2

c = 0

We've successfully determined the constants: a = 1, b = 1, and c = 0. Now, we can write the general term of the sequence:

An = n^2 + n

This is the formula that generates our sequence! To find the next terms, we can simply plug in the corresponding values of n. For example, to find the 6th term (A6), we would do:

A6 = 6^2 + 6 = 36 + 6 = 42

So, the next term in the sequence is 42. We've successfully decoded the numerical sequence and can now predict future terms. High five, mathletes!

Wrapping Up

Today, we've tackled two exciting math challenges. We conquered the rectangular square's perimeter puzzle using algebraic equations and skillfully deciphered a numerical sequence by identifying its quadratic nature and deriving its general term. Remember, math is like a puzzle – sometimes it takes a bit of digging and pattern recognition to find the solution, but the feeling of accomplishment is always worth it.

Keep those brains sharp, and happy problem-solving!