Generate Sequences With A Recursive Formula

by Andrew McMorgan 44 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of mathematics, specifically tackling a super common problem type: figuring out which sequence fits a given recursive formula. It sounds a bit technical, but trust me, it's totally doable and even kinda fun once you get the hang of it. We've got this tricky recursive formula here: f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n). Our mission, should we choose to accept it, is to find which of the given sequences could possibly be generated using this rule. Let's break it down, step by step, and make sure we nail this. Get ready to flex those math muscles, because we're about to unravel this puzzle together!

Understanding the Recursive Formula

So, what exactly is a recursive formula? In simple terms, it's a way to define a sequence where each term is calculated based on the previous term. Our formula, f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n), is a prime example. It tells us that to get the next term in the sequence (that's f(n+1)f(n+1)), you need to take the current term (that's f(n)f(n)) and multiply it by 1.5. This multiplier, 1.5, is called the common ratio because it's the constant value you multiply by to get from one term to the next. This type of sequence is technically known as a geometric sequence. The key takeaway here is that each term is 1.5 times the term before it. This is the golden rule we need to keep in mind as we check our options. We're not just looking for any pattern; we're looking for a specific pattern of multiplying by 1.5. So, when you see f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n), think of it as 'next term equals current term times 1.5'. This simple rule is our compass, guiding us through the options.

Analyzing the Options

Now, let's put our detective hats on and examine each of the sequences provided. Remember, we're looking for the one sequence where every single term is exactly 1.5 times the term before it. If we find even one pair of consecutive terms that doesn't follow this rule, we can confidently cross that option off our list. We'll start with the first term of each sequence and see if multiplying it by 1.5 gives us the second term, then check if the second multiplied by 1.5 gives us the third, and so on. This systematic approach ensures we don't miss anything.

Option A: -12, -18, -27, ...

Let's start with sequence A. The first term is -12. According to our formula, the next term should be f(2)=1.5imesf(1)f(2) = 1.5 imes f(1). So, we calculate 1.5imes(βˆ’12)1.5 imes (-12).

1.5imes(βˆ’12)=βˆ’181.5 imes (-12) = -18

Great! The second term in option A is indeed -18. So far, so good. Now, let's check the step from the second term (-18) to the third term (-27). We need to see if f(3)=1.5imesf(2)f(3) = 1.5 imes f(2).

1.5imes(βˆ’18)=βˆ’271.5 imes (-18) = -27

Bingo! The third term is also -27. This looks very promising, guys! It seems like sequence A perfectly matches our recursive formula. Every term is 1.5 times the previous term. This is exactly what the formula f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n) dictates. The ratio between consecutive terms is consistently 1.5. This consistency is the hallmark of a geometric sequence defined by this specific recursive relation. It’s important to remember that the initial value, f(1)f(1), can be anything, but once that first term is set, the rest of the sequence is determined by the multiplication factor. In this case, starting with -12, and applying the rule of multiplying by 1.5, leads directly to -18, then -27, and so on. The negative signs are also consistent; multiplying a negative number by a positive number (1.5) always results in a negative number, which is maintained throughout this sequence.

Option B: -20, 30, -45, --

Let's move on to sequence B: -20, 30, -45, ... The first term is -20. Let's apply our rule: f(2)=1.5imesf(1)f(2) = 1.5 imes f(1).

1.5imes(βˆ’20)=βˆ’301.5 imes (-20) = -30

Wait a minute! The second term in sequence B is 30, not -30. This immediately tells us that sequence B does not follow the formula f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n). The sign is wrong, and even if the absolute values matched, the sign difference disqualifies it. The formula specifies a multiplication by a positive 1.5, which means the sign of the terms should remain the same if the first term is positive, or remain negative if the first term is negative. Here, the sign flips from negative to positive, which would require multiplication by a negative ratio. However, our ratio is 1.5, which is positive. This mismatch is a clear indicator that this sequence is not generated by the given formula. We can stop checking this option right here; it's a dead end for us. It's crucial to pay attention to both the magnitude and the sign when working with these formulas. A single discrepancy is enough to rule out an option.

Option C: -18, -16.5, -15, --

Now, let's examine sequence C: -18, -16.5, -15, ... The first term is -18. Let's apply our formula: f(2)=1.5imesf(1)f(2) = 1.5 imes f(1).

1.5imes(βˆ’18)=βˆ’271.5 imes (-18) = -27

Uh oh. The second term in sequence C is -16.5, not -27. This sequence also fails the test right at the second term. What’s interesting here is that the difference between -18 and -16.5 is -1.5 (-16.5 - (-18) = 1.5), and the difference between -16.5 and -15 is also 1.5 (-15 - (-16.5) = 1.5). This indicates that sequence C is an arithmetic sequence with a common difference of 1.5 (or -1.5 if we look at it as f(n+1)=f(n)βˆ’1.5f(n+1) = f(n) - 1.5), not a geometric sequence with a common ratio of 1.5. The formula f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n) requires multiplication, not addition or subtraction. Therefore, sequence C is definitely not the one we're looking for. It’s a common trap to confuse arithmetic and geometric sequences, so always double-check whether the operation is multiplication (ratio) or addition/subtraction (difference).

Option D: -16, -17.5, -19, ...

Finally, let's check out sequence D: -16, -17.5, -19, ... The first term is -16. Applying our formula: f(2)=1.5imesf(1)f(2) = 1.5 imes f(1).

1.5imes(βˆ’16)=βˆ’241.5 imes (-16) = -24

Again, we have a mismatch. The second term in sequence D is -17.5, not -24. Similar to sequence C, let's look at the differences. The difference between -16 and -17.5 is -1.5 (-17.5 - (-16) = -1.5). The difference between -17.5 and -19 is also -1.5 (-19 - (-17.5) = -1.5). This confirms that sequence D is also an arithmetic sequence with a common difference of -1.5. It's not a geometric sequence generated by multiplying by 1.5. So, sequence D is also ruled out. It's good that we checked all the options thoroughly. Sometimes, you might find multiple options that seem close, but only one will perfectly fit the given recursive rule. The key is strict adherence to the defined operation and ratio.

Conclusion

After carefully analyzing each option against the recursive formula f(n+1)=1.5imesf(n)f(n+1) = 1.5 imes f(n), we found that only Option A: -12, -18, -27, ... consistently follows the rule. In this sequence, each term is indeed 1.5 times the previous term: βˆ’12imes1.5=βˆ’18-12 imes 1.5 = -18, and βˆ’18imes1.5=βˆ’27-18 imes 1.5 = -27. The other options failed to meet this condition, with some being arithmetic sequences or having incorrect signs. So, the sequence that could be generated using the formula is Option A. Pretty straightforward once you break it down, right? Keep practicing these, and you'll be a pro at spotting recursive sequences in no time! Stay tuned for more math adventures here at Plastik Magazine!