Which Term of AP 121, 117, 113 is Its First Negative Term? Finding the Threshold

Unlocking the Mystery: Which Term of AP 121, 117, 113 is Its First Negative Term?

I remember grappling with arithmetic progression problems back in high school. There was this one particular question that stuck with me: “Which term of AP 121, 117, 113 is its first negative term?” It seemed straightforward enough at first glance, but as I delved into it, I realized it required more than just a quick guess. It demanded a solid understanding of the underlying mathematical principles. My teacher, Mrs. Davison, a wonderfully patient woman with a knack for making math accessible, guided us through the process. She’d often say, "Think of it like a descending staircase, folks. Each step takes you lower, and eventually, you're going to hit the basement – the negative numbers!" That analogy, simple as it was, really helped me visualize the sequence and the inevitable descent into negativity. Now, years later, I find myself revisiting this classic problem, and I want to share the in-depth analysis and clear methodology that will not only answer the question for the sequence 121, 117, 113 but also equip you with the skills to tackle any similar arithmetic progression challenge.

So, to directly answer the question: The 31st term of the arithmetic progression 121, 117, 113 is its first negative term. This might seem like a precise answer without much fanfare, but the journey to arrive at this number involves understanding the fundamental properties of arithmetic progressions and applying a systematic approach. Let's break down exactly how we get there.

Understanding Arithmetic Progressions (APs)

Before we dive into finding the first negative term, it’s crucial to have a firm grasp on what an arithmetic progression is. In simple terms, an AP is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is known as the "common difference." Think of it as a steady beat in a rhythm, or a consistent stride as you walk. Each number in the sequence is obtained by adding this common difference to the previous one.

Key Components of an AP

Every arithmetic progression has two primary components that define it:

The First Term (a): This is simply the starting number of the sequence. In our specific problem, the first term is 121. The Common Difference (d): This is the constant value added to each term to get the next. It’s the 'step size' of our progression.

To find the common difference (d) in any AP, you just subtract any term from its succeeding term. For example, using our sequence 121, 117, 113:

117 - 121 = -4 113 - 117 = -4

So, the common difference (d) for our AP is -4. This negative value tells us that the terms are decreasing, which is precisely why we expect to eventually find negative terms.

The General Term Formula

A powerful tool in arithmetic progressions is the formula for the nth term. This formula allows us to calculate any term in the sequence without having to list out all the preceding terms. It's a shortcut that saves a lot of time and effort, especially for sequences with many terms.

The formula for the nth term of an AP is:

$a_n = a + (n-1)d$

Where:

$a_n$ is the nth term (the term we want to find). $a$ is the first term. $n$ is the term number (e.g., 1st, 2nd, 3rd, etc.). $d$ is the common difference.

Let's quickly test this formula with our sequence. We know $a = 121$ and $d = -4$.

For the 1st term (n=1): $a_1 = 121 + (1-1)(-4) = 121 + 0 = 121$. Correct! For the 2nd term (n=2): $a_2 = 121 + (2-1)(-4) = 121 + 1(-4) = 121 - 4 = 117$. Correct! For the 3rd term (n=3): $a_3 = 121 + (3-1)(-4) = 121 + 2(-4) = 121 - 8 = 113$. Correct!

As you can see, the formula accurately represents the sequence.

The Quest for the First Negative Term

Now, the core of our problem: finding the first term that is negative. A negative term is any term with a value less than zero. Using our general term formula, we want to find the smallest integer 'n' (since 'n' represents the term number, it must be a positive integer) for which $a_n < 0$.

So, we set up the inequality:

$a_n < 0$

Substitute the general term formula into this inequality:

$a + (n-1)d < 0$

Now, let's plug in the values we know for our specific AP: $a = 121$ and $d = -4$.

$121 + (n-1)(-4) < 0$

Solving the Inequality: A Step-by-Step Approach

This inequality is the key to unlocking our answer. We need to solve for 'n' while remembering that 'n' must be a positive whole number. Here's how we can approach it:

Distribute the common difference:

$121 - 4(n-1) < 0$

$121 - 4n + 4 < 0$

Combine constant terms:

$125 - 4n < 0$

Isolate the term with 'n': To do this, we can add $4n$ to both sides of the inequality. This is a standard algebraic manipulation.

$125 < 4n$

Solve for 'n': Now, divide both sides by 4. It's important to remember that when you divide an inequality by a positive number, the direction of the inequality sign remains the same.

$125 / 4 < n$

$31.25 < n$

So, the inequality $a_n < 0$ is true when $n > 31.25$.

Interpreting the Result

What does $n > 31.25$ mean in the context of our arithmetic progression? Remember, 'n' represents the term number, and term numbers must be whole numbers (1, 2, 3, and so on). We are looking for the *smallest integer* 'n' that satisfies this condition.

If $n$ must be greater than 31.25, the very next whole number after 31.25 is 32. However, let's pause and think carefully. The inequality $a_n < 0$ means the term is *strictly less than zero*.

Consider the terms around $n = 31.25$.

If $n = 31$, then $n$ is *not* greater than 31.25. So, the 31st term is likely not negative, or it might be zero (though unlikely given our calculations). If $n = 32$, then $n$ *is* greater than 31.25. So, the 32nd term is definitely negative.

The inequality $n > 31.25$ tells us that any term *after* the point where the value crosses zero will be negative. The number 31.25 represents the theoretical point where the sequence *would* hit exactly zero if fractional terms were allowed. Since we can only have whole-numbered terms, we need to find the first whole number 'n' that makes the term negative.

The inequality $n > 31.25$ means the smallest integer value for $n$ that satisfies this is 32. This implies that the 32nd term is the first one that is strictly less than zero. Let's double-check this by calculating the 31st and 32nd terms.

Calculating the 31st Term ($a_{31}$)

$a_{31} = a + (31-1)d$

$a_{31} = 121 + (30)(-4)$

$a_{31} = 121 - 120$

$a_{31} = 1$

So, the 31st term is 1, which is positive.

Calculating the 32nd Term ($a_{32}$)

$a_{32} = a + (32-1)d$

$a_{32} = 121 + (31)(-4)$

$a_{32} = 121 - 124$

$a_{32} = -3$

The 32nd term is -3, which is indeed negative.

This confirms that the 31st term is positive (1), and the very next term, the 32nd term, is negative (-3). Therefore, the 32nd term is the first negative term in the sequence.

Refining the Inequality Approach: A Subtle but Important Point

My initial interpretation of $n > 31.25$ leading directly to $n=32$ as the first term *less than* zero is correct. However, sometimes, depending on how the inequality is set up or solved, one might arrive at a slightly different phrasing that requires careful consideration. Let's revisit the inequality $125 - 4n < 0$.

If we decided to isolate the $4n$ term differently:

$125 - 4n < 0$

Subtract 125 from both sides:

$-4n < -125$

Now, to solve for 'n', we must divide both sides by -4. **Crucially, when you divide or multiply an inequality by a negative number, you must reverse the direction of the inequality sign.**

$n > (-125) / (-4)$

$n > 31.25$

This leads to the same result, $n > 31.25$. The smallest integer greater than 31.25 is 32. This indicates that the 32nd term ($a_{32}$) is the first term that satisfies the condition $a_n < 0$.

Let's consider another way to frame the problem. What if we want to find the *last positive term*? A positive term is any term $a_n > 0$.

$a + (n-1)d > 0$

$121 + (n-1)(-4) > 0$

$121 - 4n + 4 > 0$

$125 - 4n > 0$

Add $4n$ to both sides:

$125 > 4n$

Divide by 4:

$125 / 4 > n$

$31.25 > n$

This inequality, $31.25 > n$, tells us that the term number 'n' must be less than 31.25 for the term to be positive. The largest integer value for 'n' that is less than 31.25 is 31.

This means that the 31st term is the last term that is positive. If the 31st term is the last positive term, then the very next term, the 32nd term, must be the first one that is not positive. Since our common difference is negative, this "not positive" term will be negative.

This alternative framing of finding the last positive term and then moving to the next term often feels more intuitive for many learners, and it yields the same conclusive answer. Both methods are mathematically sound and demonstrate a deep understanding of how to manipulate arithmetic progressions.

Visualizing the Descent

Sometimes, a visual aid can solidify understanding. Imagine plotting the terms of this arithmetic progression on a graph, with the term number 'n' on the x-axis and the term value $a_n$ on the y-axis. Since the common difference is constant, the points would form a straight line. The first term is at (1, 121), the second at (2, 117), and so on.

As the common difference is negative ($d=-4$), this line will slope downwards. We are looking for the point where this line crosses the x-axis (where $a_n = 0$) and then continues into the negative region of the y-axis. Our calculation showed that the line would theoretically cross zero at $n=31.25$.

The graph would look something like this conceptually:

At n=31, the y-value is positive (1). At n=31.25, the y-value is zero. At n=32, the y-value is negative (-3).

Since we can only have integer term numbers, the 31st term is the last one above zero, and the 32nd term is the first one below zero.

Common Pitfalls and How to Avoid Them

When working with inequalities and arithmetic progressions, a few common mistakes can creep in:

Forgetting to reverse the inequality sign: As demonstrated, dividing by a negative number requires flipping the sign. This is a frequent oversight that can lead to an incorrect answer. Always be mindful of the sign of the number you're dividing or multiplying by. Rounding errors: While our calculation $125/4$ resulted in a clean 31.25, some calculations might yield repeating decimals. It's best to keep fractions or use a sufficient number of decimal places to maintain accuracy until the final step of determining the integer term number. Misinterpreting the inequality: Does $n > 31.25$ mean 31 or 32 is the first term? Always remember that 'n' must be an integer, and you're looking for the smallest integer that satisfies the condition. If $n > 31.25$, the smallest such integer is 32. If the inequality were $n < 31.25$, the largest such integer would be 31. Assuming the first term might be zero or negative: While in this specific problem, the common difference is negative, it's possible for an AP to start with negative terms or have a common difference of zero. Always verify the first few terms and the common difference to understand the sequence's behavior.

By systematically applying the general term formula and carefully solving the resulting inequality, while paying close attention to the rules of inequality manipulation, we can confidently determine the first negative term of any arithmetic progression.

Let's Practice with Another Example

To solidify the understanding, let's tackle another problem: In the AP 50, 47, 44, ..., which term is the first negative term?

Step 1: Identify the first term (a) and the common difference (d).

$a = 50$ $d = 47 - 50 = -3$

Step 2: Set up the inequality for the nth term to be negative ($a_n < 0$).

$a + (n-1)d < 0$

$50 + (n-1)(-3) < 0$

Step 3: Solve the inequality.

$50 - 3(n-1) < 0$

$50 - 3n + 3 < 0$

$53 - 3n < 0$

$53 < 3n$

$53/3 < n$

$17.666... < n$

Step 4: Interpret the result.

We need the smallest integer 'n' that is greater than 17.666.... This integer is 18.

Therefore, the 18th term of the AP 50, 47, 44, ... is its first negative term.

Let's verify:

17th term: $a_{17} = 50 + (17-1)(-3) = 50 + 16(-3) = 50 - 48 = 2$ (Positive) 18th term: $a_{18} = 50 + (18-1)(-3) = 50 + 17(-3) = 50 - 51 = -1$ (Negative)

The calculation is correct!

Why is Understanding This Important?

While finding the first negative term of a specific AP might seem like an academic exercise, the underlying principles are broadly applicable. This type of problem hones your skills in:

Algebraic manipulation: Working with formulas and solving inequalities. Logical reasoning: Interpreting mathematical results in a real-world context (even if the context is mathematical). Pattern recognition: Identifying the predictable nature of arithmetic sequences. Problem-solving strategy: Breaking down a complex problem into smaller, manageable steps.

These skills are invaluable in various fields, from finance and economics (where sequences and trends are analyzed) to computer science and engineering (where algorithms often rely on iterative processes and predictable patterns).

Frequently Asked Questions (FAQs)

Q1: How do I know if an arithmetic progression will eventually have negative terms?

You can determine this by looking at the common difference, 'd'.

If the common difference ($d$) is negative, and the first term ($a$) is positive, then the terms will decrease. Eventually, they will cross zero and become negative. For example, our original AP 121, 117, 113 has $a=121$ (positive) and $d=-4$ (negative), so it's guaranteed to have negative terms.

If the common difference ($d$) is positive, and the first term ($a$) is negative, then the terms will increase. Eventually, they will cross zero and become positive. For example, an AP like -10, -8, -6... will eventually have positive terms.

If both the first term ($a$) and the common difference ($d$) are negative, all terms will be negative. For instance, -5, -10, -15...

If both the first term ($a$) and the common difference ($d$) are positive, all terms will be positive. For instance, 2, 4, 6...

Therefore, for an AP to have negative terms, either the first term must be negative and the common difference positive (so it climbs into positives and then perhaps we might be looking for the first term *after* it crosses a certain positive value), or, more commonly as in our problem, the first term is positive and the common difference is negative.

Q2: What if the common difference is zero?

If the common difference ($d$) is zero, then every term in the sequence is the same as the first term ($a$). For example, the sequence 7, 7, 7, 7... has $a=7$ and $d=0$. In such a case, if the first term is positive, all terms will be positive. If the first term is negative, all terms will be negative. If the first term is zero, all terms will be zero. An AP with a common difference of zero will never change its sign, so it will never transition from positive to negative or vice versa unless it starts at zero.

Q3: Can an arithmetic progression have a term that is exactly zero?

Yes, it's possible for a term in an arithmetic progression to be exactly zero. This happens if the value of 'n' that solves the equation $a + (n-1)d = 0$ is an integer.

Let's consider an example. Suppose we have the AP: 10, 7, 4, 1, -2, ...

Here, $a = 10$ and $d = -3$.

We want to find if there's a term where $a_n = 0$:

$10 + (n-1)(-3) = 0$

$10 - 3n + 3 = 0$

$13 - 3n = 0$

$13 = 3n$

$n = 13/3$

Since $n = 13/3$ is not an integer, this specific AP will never have a term that is exactly zero. It will jump from positive values (like 1) to negative values (like -2).

Now, consider an AP where a term *is* zero. For instance, an AP where $a=6$ and $d=-2$.

$6 + (n-1)(-2) = 0$

$6 - 2n + 2 = 0$

$8 - 2n = 0$

$8 = 2n$

$n = 4$

This means the 4th term is 0. Let's check: $a_1=6, a_2=4, a_3=2, a_4=0$. Indeed, the 4th term is zero. In this case, the "first negative term" would be the term immediately following the zero term, which would be $a_5 = 0 + (-2) = -2$. So, the 5th term would be the first negative term.

Q4: How does the number of terms affect finding the first negative term?

The number of terms in an arithmetic progression is not a fixed limit when we are asked to find a specific type of term, like the first negative term. We are essentially assuming the progression continues indefinitely in the same pattern. The question is about the *nature* of the sequence as it progresses, not about a finite set of terms.

If the question were phrased like "In the first 20 terms of AP X, is there a negative term?", then we would calculate the 20th term and check its sign, and also check if any terms *before* the 20th were negative. However, the phrasing "Which term... is its first negative term" implies we are looking for the index 'n' where the sequence *first* dips below zero, assuming it continues.

So, even if an AP had only, say, 10 terms listed, but the pattern suggests it would go negative later, we'd still calculate based on the infinite continuation of the pattern. Our calculation for the AP 121, 117, 113 determined that the 32nd term is the first negative one. This means if the sequence were to continue, the 32nd term would be the first to exhibit a negative value.

Q5: I keep getting fractional answers for 'n'. What am I doing wrong?

You are likely not doing anything "wrong" in the calculation itself, but rather misinterpreting the result. As we've seen, the value of 'n' derived from inequalities like $n > 31.25$ or $n < 31.25$ often results in a decimal or fraction. This is perfectly normal!

The key is that 'n' must represent a term *number*, which must be a positive integer. The decimal result from the inequality calculation represents the exact mathematical point where the AP would cross zero (if fractional terms were allowed). You then need to find the *closest integer* to this point that satisfies the inequality.

For example:

If your inequality results in $n > \text{some decimal}$, you need to find the smallest integer *greater than* that decimal. That would be the integer immediately following the decimal value. If your inequality results in $n < \text{some decimal}$, you need to find the largest integer *less than* that decimal. That would be the integer immediately preceding the decimal value.

It's about finding the *first whole number* that fits the condition, based on the mathematical threshold indicated by the fractional result.

Conclusion: Mastering the First Negative Term

We've thoroughly explored the arithmetic progression 121, 117, 113, and by applying the principles of arithmetic progressions and solving inequalities, we've definitively identified that the 32nd term is its first negative term. This journey involved understanding the fundamental definition of an AP, utilizing the general term formula, setting up and solving inequalities, and carefully interpreting the results in the context of integer term numbers.

This process, while specific to this particular sequence, is a robust method applicable to any arithmetic progression. By mastering these steps, you gain a powerful tool for analyzing sequences and predicting their behavior. Whether you're a student facing a textbook problem or a professional analyzing trends, the ability to systematically find such thresholds is incredibly valuable. Remember Mrs. Davison's staircase analogy; with each step down (negative common difference), you're bound to reach the negative numbers eventually. Our task was simply to find out precisely *when* that descent truly begins.

The sequence starts at a positive value and decreases consistently. The question of which term is the first negative term is essentially asking: at what point does this consistent decrease result in a value less than zero? By using the formula $a_n = a + (n-1)d$ and setting $a_n < 0$, we establish an inequality that, when solved for $n$, tells us the range of term numbers that will yield negative results. In our case, $n > 31.25$. Since $n$ must be an integer, the smallest integer greater than 31.25 is 32. This confirms that the 32nd term is indeed the first one to fall into negative territory.

This problem highlights the elegance of mathematics: a seemingly simple sequence holds within it a predictable pattern that can be uncovered with the right tools and understanding. The constant difference of -4 ensures a steady decline, and our algebraic approach allows us to pinpoint the exact moment this decline crosses the zero threshold and enters the realm of negative numbers.