Solving Arithmetic Progressions for Three Terms

Unlocking Arithmetic Progressions: Solving for Three AP Terms

You've stumbled upon a classic math problem, one that's a staple of high school algebra: An Arithmetic Progression (AP) with the first three terms summing to 6 and their product equaling 6. Let's dive into the world of APs, sharpen your pencil, and discover the three elusive numbers.

An AP is a sequence where the difference between consecutive terms is constant. In general form, an AP can be written as (a_n = a_1 + (n-1)d), where (a_1) is the first term, (d) is the common difference, and (n) is the term number.

The Problem

To find the three terms, let's call them (a_1), (a_2), and (a_3), we're given:

1. The sum of the first three terms: (\sum_{n=1}^3 a_n = a_1 + a_2 + a_3 = 6)
2. The product of the first three terms: (a_1 \times a_2 \times a_3 = 6).

Solving the Problem

Let's first find the common difference (d). Since (a_2) is two terms ahead of (a_1), (d = a_2 - a_1). From the first equation:

[a_1 + a_2 + a_3 = 6]

Substituting (a_2 = a_1 + d):

[a_1 + (a_1 + d) + a_3 = 6]

Combining like terms:

[2a_1 + a_3 = 6]

Now let's find (a_3). From the second equation:

[a_1 \times a_2 \times a_3 = 6]

Substituting (a_2 = a_1 + d):

[a_1 \times (a_1 + d) \times a_3 = 6]

Expanding:

[a_1^2 + a_1d + da_3 = 6]

But we already have an equation for (a_3), so substitute (2a_1 + a_3 = 6):

[a_1^2 + a_1d + 2a_1 - 6 = 6]

Combining like terms:

[a_1^2 + 3a_1 - 6 = 0]

This is a quadratic equation, which we can solve by factoring or the quadratic formula:

[a_1 = \frac{-3 \pm \sqrt{(3)^2 - 4 \times 1 \times (-6)}}{2 \times 1}]

[a_1 = \frac{-3 \pm \sqrt{9 + 24}}{2}]

[a_1 = \frac{-3 \pm \sqrt{33}}{2}]

Let's call these two solutions (a_{11}) and (a_{12}). Now that we have (a_1), we can find (a_2 = a_1 + d) and (a_3 = 2a_1 + 6 - a_2). Given these values, we can check which one satisfies the problem statement.

The Answer

The easiest way to check is by plugging the first three terms back into the original equations. For the sum equation, using one of the solutions:

[a_{11} + (a_{11} + d) + a_3 = 6]

Substitute (a_3 = 2a_{11} + 6 - (a_{11} + d)):

[a_{11} + a_{11} + d + 2a_{11} + 6 - (a_{11} + d) = 6]

Combining like terms:

[3a_{11} + 6 = 6]

[a_{11} = \frac{6 - 6}{3}]

[a_{11} = 0]

Now find (a_2) and (a_3):

[a_2 = a_{11} + d]

[a_3 = 2a_{11} + 6 - a_2]

Plugging in (a_{11} = 0):

[a_2 = 0 + d \rightarrow d = a_2]

[a_3 = 2 \times 0 + 6 - (0 + d) \rightarrow a_3 = 6 - d]

Now check the product equation:

[a_{11} \times a_2 \times a_3 = 6]

[0 \times a_2 \times (6 - a_2) = 6]

At (a_2 = d = 0), the product equation is satisfied. Therefore, the three AP terms that satisfy the problem are (a_1 = 0), (a_2 = 0), and (a_3 = 6).

So there you have it—the solution to the AP problem of finding three terms with a sum of 6 and product of 6. Happy computing!