Question

Find the 15 th term of the arithmetic progression \(10,4,-2, \ldots\) (1) \(-72\) (2) \(-74\) (3) \(-76\) (4) \(-78\)

Short answer

Question: Calculate the 15th term of the arithmetic progression, if the first three terms are 10, 4, and -2. Answer: (2) -74

Step-by-step solution

Find the common difference

To find the common difference, we can subtract any consecutive terms in the AP, such as the second term minus the first term, or the third term minus the second term. We calculate: \(4-10 = -6\) and \(-2-4=-6\). So, the common difference is -6.

Apply the nth term formula

Now, we will apply the nth term formula: \(T_n = a + (n-1) \times d\), where \(T_{15}\) is the 15th term, \(a = 10\), \(n = 15\), and \(d = -6\).

Substitute the values and solve for the 15th term

We will plug in the values \(a = 10\), \(n = 15\) and \(d = -6\) into the formula for the nth term: \(T_{15} = 10 + (15-1) \times (-6)\) \(T_{15} = 10 + (14) \times (-6)\) \(T_{15} = 10 - 84\) \(T_{15} = -74\)

Identify the answer in the given choices

We have found the 15th term of the arithmetic progression to be \(-74\). Comparing this with the given options, we see that the correct answer is option (2) \(-74\).

Key concepts

Common Difference

When you encounter an arithmetic progression, one of the foundational elements is the common difference. It is the consistent amount you add (or subtract, if it's negative) to each term to get to the next. To find the common difference, just subtract any term in the sequence from the term that follows it.
In this exercise, we took the second term, 4, and subtracted the first term, 10, resulting in \( 4 - 10 = -6 \). We confirmed it by checking the difference between the third term, -2, and the second term, 4, which also gave us \(-2 - 4 = -6\).

• This consistency shows that the common difference is -6.
• The common difference is vital for calculating other terms in the sequence.

Understanding and identifying the common difference allows you to dive deeper into arithmetic progressions and solve more complex problems with confidence.

Nth Term Formula

The nth term formula is your key to finding any term in an arithmetic progression without listing all previous ones. It's a simple and powerful tool. Given by the equation:\[T_n = a + (n-1) \times d\]where:

• \(T_n\) is the nth term you're looking for.
• \(a\) is the first term of the sequence.
• \(n\) is the position of the term in the sequence.
• \(d\) is the common difference.

To find the 15th term in our arithmetic progression:

• First, we identify \(a\) as 10 (the first term), \(n\) as 15, and \(d\) as -6.
• Then apply these values into the formula.

Using this formula makes finding specific terms much more efficient, especially for larger n.

Sequence Calculation

Armed with the formula, sequence calculation becomes straightforward. By plugging the values into our formula, we calculate the specific term we need. Calculating the 15th term for our sequence follows these steps:

• Start with the nth term formula: \(T_n = a + (n-1) \times d\).
• Substitute \(a = 10\), \(n = 15\), and \(d = -6\) into the formula.

This looks like: \[T_{15} = 10 + (15-1) \times (-6)\]Simplify by computing the results step by step:

• Calculate the expression in parentheses, \(15 - 1 = 14\).
• Multiply by the common difference: \(14 \times -6 = -84\).
• Then add this to the first term: \(10 - 84 = -74\).

Thus, the 15th term is \(-74\). This calculation shows the methodical way of finding any sequence term.

Mathematical Problem-Solving

In mathematical problem-solving, having a systematic approach like the one for arithmetic sequences is crucial. Start by identifying all the components of the problem:

• Recognize what you know: the first term \(a\) and the position \(n\) you want.
• Find the unknown: the common difference \(d\) and use it to resolve the sequence.

With all data collected, apply the nth term formula to find the desired term. This structured process not only provides solutions but also builds your mathematical thinking and strategy skills. Remember, solving these problems enhances your analytical capabilities, helping you to tackle both academic and real-world challenges effectively.Whether dealing with a simple arithmetic progression or more complicated sequences, breaking down the problem into easily digestible steps is key. Practice often to improve your problem-solving efficiency and confidence.