AP Sequence: Find The 5th Term With Given Conditions

by TextBrain Team 53 views

Hey guys! Let's dive into a cool math problem involving arithmetic progressions (APs). We've got a sequence of 6 numbers in AP, and we need to figure out the 5th term. The problem gives us two key pieces of information: the sum of all 6 numbers is 3, and the first term is 4 times the third term. Sounds like a fun challenge, right? So, let's break it down and solve it step by step!

Understanding Arithmetic Progressions (APs)

Before we jump into the specifics of this problem, let's quickly recap what an arithmetic progression is. An AP is simply a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference, often denoted as 'd'.

Think of it like this: you start with a number (the first term, 'a') and then keep adding the same number ('d') to get the next term. So, the sequence looks like: a, a + d, a + 2d, a + 3d, and so on.

Key terms to remember:

  • a: The first term of the AP.
  • d: The common difference.
  • n: The number of terms in the AP.
  • an: The nth term of the AP (given by the formula an = a + (n-1)d).
  • Sn: The sum of the first n terms of the AP (given by the formula Sn = n/2 * [2a + (n-1)d]).

Now that we've refreshed our understanding of APs, let's get back to our problem!

Setting up the Equations

Okay, so we know we have 6 numbers in AP. Let's represent these terms using our AP knowledge:

  1. First term: a
  2. Second term: a + d
  3. Third term: a + 2d
  4. Fourth term: a + 3d
  5. Fifth term: a + 4d
  6. Sixth term: a + 5d

Remember, a is the first term, and d is the common difference. The problem gives us two crucial pieces of information that we can turn into equations:

  1. The sum of all 6 numbers is 3: This means a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d) = 3
  2. The first term is 4 times the third term: This translates to a = 4(a + 2d)

Now we have two equations with two unknowns (a and d). This is fantastic because we can solve for these variables!

Let's simplify these equations a bit.

Simplifying the Sum Equation

Let's tackle the first equation, the sum of the terms:

a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d) = 3

Combine all the 'a' terms and all the 'd' terms:

6a + (d + 2d + 3d + 4d + 5d) = 3

6a + 15d = 3

We can simplify this further by dividing the entire equation by 3:

2a + 5d = 1

Awesome! We've got our first simplified equation: 2a + 5d = 1.

Simplifying the Term Relationship Equation

Now let's simplify the second equation, which relates the first term to the third term:

a = 4(a + 2d)

Distribute the 4 on the right side:

a = 4a + 8d

Subtract 'a' from both sides:

0 = 3a + 8d

So, our second simplified equation is: 3a + 8d = 0.

Solving the System of Equations

Alright, we've got a system of two linear equations:

  1. 2a + 5d = 1
  2. 3a + 8d = 0

There are a couple of ways we can solve this system. We can use substitution or elimination. Let's use the elimination method, which often works well for these types of problems.

Using Elimination Method

To use elimination, we want to multiply the equations by constants so that either the 'a' coefficients or the 'd' coefficients are opposites. Let's eliminate 'a'.

Multiply the first equation by 3:

3 * (2a + 5d) = 3 * 1 => 6a + 15d = 3

Multiply the second equation by -2:

-2 * (3a + 8d) = -2 * 0 => -6a - 16d = 0

Now, add the two modified equations together:

(6a + 15d) + (-6a - 16d) = 3 + 0

The '6a' and '-6a' cancel out, leaving us with:

-d = 3

So, d = -3. We've found the common difference!

Finding the First Term (a)

Now that we know d, we can plug it back into either of our simplified equations to solve for a. Let's use the equation 3a + 8d = 0:

3a + 8(-3) = 0

3a - 24 = 0

3a = 24

a = 8. We've found the first term!

Calculating the 5th Term

We're almost there! The question asked us to find the 5th term of the AP. Remember the formula for the nth term:

an = a + (n-1)d

For the 5th term (n = 5), we have:

a5 = a + (5-1)d

a5 = a + 4d

Now, plug in the values we found for a and d:

a5 = 8 + 4(-3)

a5 = 8 - 12

a5 = -4

Final Answer

So, the 5th term of the arithmetic progression is -4.

Therefore, the 5th term of the AP is -4.

Key Takeaways

This problem was a great example of how to use the properties of arithmetic progressions to solve a problem. Here are the key takeaways:

  • Understand the definition of an AP and its key terms (a, d, n, an, Sn).
  • Be able to translate word problems into mathematical equations.
  • Know how to solve systems of linear equations (substitution or elimination).
  • Remember the formula for the nth term of an AP: an = a + (n-1)d.

I hope you guys found this explanation helpful! Let me know if you have any questions or want to tackle another math problem together. Keep practicing, and you'll become a math whiz in no time!