Finding The Smallest Term In A Sequence

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Finding the Smallest Term in a Sequence: A Detailed Guide

Hey guys! Let's dive into a common algebra problem: finding the smallest term (the minimum value) within a sequence. This is a super handy skill, and we'll break it down step by step. We'll be looking at three different sequences defined by formulas, so you'll get plenty of practice. Ready to get started? Let's go!

1. Analyzing the First Sequence: a_n = n² - 12n

Alright, let's tackle the first sequence! We have a_n = n² - 12n. Our goal is to figure out the smallest possible value this sequence can produce. One cool way to approach this is by recognizing that this formula represents a parabola. The n² term tells us it's a parabola that opens upwards (like a smile!), meaning it has a minimum point (a vertex). To find this minimum, we can use a couple of methods. First, we could complete the square. This technique transforms the quadratic expression into a form that immediately reveals the vertex. Let's do it! We'll rewrite n² - 12n to complete the square. We take half of the coefficient of our n term (-12), square it ((-12/2)² = 36), and add and subtract it within the expression: n² - 12n + 36 - 36. Now we can rewrite the first three terms as a perfect square: (n - 6)² - 36. This is our completed square form! From this, we can easily see the vertex of the parabola is at the point (6, -36). This means the minimum value of the sequence occurs when n = 6, and the smallest term is -36. Another method, and perhaps a quicker one, is using the vertex formula. For a quadratic in the form of an² + bn + c, the n-coordinate of the vertex (where the minimum or maximum occurs) is given by n = -b / 2a. In our case, a = 1, b = -12, and c = 0. So, n = -(-12) / (2 * 1) = 6. Plugging n = 6 back into our formula: a_6 = 6² - 12 * 6 = 36 - 72 = -36. Boom! Same answer. The smallest term in the sequence is -36, and it occurs at the 6th term. Pretty neat, right? Now, you can see how both methods work and get the same answer. It's about which one you find easier or faster to apply!

Now, let's explore this first sequence further. Let's calculate the values for a few terms around n = 6 to confirm our findings. When n = 5, a_5 = 5² - 12 * 5 = 25 - 60 = -35. When n = 6, as we've calculated, a_6 = -36. When n = 7, a_7 = 7² - 12 * 7 = 49 - 84 = -35. See? The values decrease as we approach n=6 and then increase again, confirming that -36 is indeed the smallest term. Also, the sequence is symmetric around the vertex, as expected. So, by understanding parabolas, completing the square, or using the vertex formula, we've successfully found the minimum value. Keep in mind that understanding these basics helps you solve more complex problems too. This is a very useful technique in various areas of mathematics, from calculus to optimization problems!

2. Investigating the Second Sequence: a_n = n² - 13n + 23

Alright, moving on to the second sequence: a_n = n² - 13n + 23. This one is slightly different, but the core idea remains the same. Again, we are dealing with a quadratic equation, meaning a parabola. To find the smallest term, we'll again aim to find the vertex of the parabola. This time, completing the square will be helpful in identifying the vertex easily. Let's rewrite the equation by completing the square: n² - 13n + 23 = n² - 13n + (13/2)² - (13/2)² + 23. We've added and subtracted (13/2)², which is 169/4. Now we rewrite the equation: (n - 13/2)² - 169/4 + 23. We can then simplify: (n - 13/2)² - 169/4 + 92/4 = (n - 13/2)² - 77/4. Therefore, the vertex of the parabola is at (13/2, -77/4). The n-coordinate of the vertex is 13/2, or 6.5. Since n represents the term number in the sequence, and term numbers must be whole numbers, we need to consider the terms on either side of n = 6.5, which are n = 6 and n = 7. Let's plug these values into the original equation: a_6 = 6² - 13 * 6 + 23 = 36 - 78 + 23 = -19. a_7 = 7² - 13 * 7 + 23 = 49 - 91 + 23 = -19. Whoa! It looks like we get the same value for both n = 6 and n = 7! This happens because the vertex falls between two integer values. The smallest term in this sequence is -19, and it occurs at both the 6th and 7th terms. Another way to approach this problem is to use the vertex formula. Recall that the n-coordinate of the vertex is given by n = -b / 2a. In this case, a = 1, b = -13, and c = 23. So, n = -(-13) / (2 * 1) = 13/2 = 6.5. This reinforces what we've already found, and now we know that the minimum term must be found in the adjacent integers. That's why we calculated the values for the 6th and 7th terms. Thus, the minimum value is -19. When you encounter situations where the vertex n-coordinate isn't a whole number, always check the nearest integers to pinpoint the smallest term. It is a very important fact that you should never forget!

Let's also visualize it: at n=5, a_5 = 5² - 135 + 23 = 25 - 65 + 23 = -17; at n=6, a_6 = -19; at n=7, a_7 = -19; at n=8, a_8 = 8² - 138 + 23 = 64 - 104 + 23 = -17. As you can see, the values drop from -17, reach a minimum of -19, and then rise again. This behavior validates our approach and demonstrates how the vertex guides us to the smallest term. Also, the symmetry around the vertex can easily be seen.

3. Examining the Third Sequence: a_n = 2n² + 5n - 3

Let's get into the third and final sequence: a_n = 2n² + 5n - 3. This one has a slightly different form, but the same principles still apply. We have a quadratic equation, so we are again dealing with a parabola. And the coefficient of n² is positive (2), meaning our parabola opens upwards and has a minimum point. Let's find this minimum value. We can complete the square again. First, factor out the 2 from the first two terms: 2(n² + (5/2)n) - 3. Now, we complete the square inside the parentheses. Take half of the coefficient of n (5/2), square it ((5/4)² = 25/16), add and subtract inside the parentheses: 2(n² + (5/2)n + 25/16 - 25/16) - 3. This becomes 2((n + 5/4)² - 25/16) - 3. Distribute the 2: 2(n + 5/4)² - 25/8 - 3. Simplify further: 2(n + 5/4)² - 25/8 - 24/8 = 2(n + 5/4)² - 49/8. This equation reveals the vertex at (-5/4, -49/8). Since the vertex is at n = -5/4 = -1.25, and n must be a positive integer in a sequence, we need to consider the terms near this vertex. Since we can't have negative terms in this situation, we look at terms n = 1, 2, 3 and so on. Also, a quadratic will not necessarily have the smallest value at n = 0, but as we are finding terms of the sequence, we need to check the closest integer terms. Let's check a_1: a_1 = 2 * 1² + 5 * 1 - 3 = 2 + 5 - 3 = 4. Checking a_2: a_2 = 2 * 2² + 5 * 2 - 3 = 8 + 10 - 3 = 15. Thus, in this case, the smallest term seems to be at n = 1. A quicker alternative, again, is the vertex formula: n = -b / 2a. Here, a = 2, b = 5, and c = -3. So n = -5 / (2 * 2) = -5/4 = -1.25, just like with completing the square. Since n must be a positive integer in the context of our sequence, we need to check the terms starting from n = 1. Therefore, when n = 1, a_1 = 4. When n = 2, a_2 = 15, and so on. Hence, the smallest term in this sequence is 4, which is the first term (a_1). This highlights the importance of understanding the context of the problem, and which methods are most appropriate. This is because we cannot use negative values for n. This approach confirms that, in a practical sense, the sequence begins at n = 1, and so we can start our search from there. And it also highlights the usefulness of knowing multiple approaches.

Let's verify our findings with a few more calculations: a_1 = 4, a_2 = 15, a_3 = 23² + 53 - 3 = 18 + 15 - 3 = 30; a_4 = 24² + 54 - 3 = 32 + 20 - 3 = 49. As you can see, the values increase rapidly, confirming that the smallest term is indeed 4. Remember, these are quadratic sequences, so they will always have that parabolic shape and symmetry. However, in this case, the context of the problem dictates that we start at n = 1, so the vertex doesn't give us the smallest possible value for the sequence.

Conclusion: Wrapping Things Up!

Alright, guys, we've explored three different sequences, and we've learned how to find their smallest terms! We used methods like completing the square and the vertex formula, which are both super valuable techniques. The key takeaway here is to recognize that these formulas represent parabolas and to find their minimum points. Also remember the context and what values of n are valid in the sequence. Keep practicing, and you'll become a pro at these types of problems! Good job, everyone! And don't hesitate to revisit these examples to reinforce your understanding. The ability to find the smallest or largest values in a sequence is incredibly useful in many branches of mathematics and real-world applications! Now go out there and conquer those sequence problems! You've got this!