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Chapter 19: Problem 19

Find the sum of the \(n\) terms of the indicated arithmetic sequence. $$-2,-\frac{5}{2},-3, \ldots ; n=10$$

Short Answer

Expert verified
The sum of the first 10 terms is \(-\frac{85}{2}\).

Step by step solution

01

Identify the first term and the common difference

The first term, denoted as \(a\), of the sequence is \(-2\). To find the common difference \(d\), subtract the first term from the second term: \(d = -\frac{5}{2} - (-2) = -\frac{5}{2} + 2 = -\frac{1}{2}\).
02

Find the formula for the nth term

The formula for the \(n\)-th term in an arithmetic sequence is given by \(a_n = a + (n-1) \cdot d\). Substituting our values, we have \(a_n = -2 + (n-1)(-\frac{1}{2})\).
03

Use the formula for the sum of the first n terms

The sum of the first \(n\) terms of an arithmetic sequence is given by \(S_n = \frac{n}{2} \cdot (a + a_n)\). First, we need to find \(a_{10}\), the 10th term. Using our formula from Step 2: \(a_{10} = -2 + (10-1)(-\frac{1}{2}) = -2 + (-\frac{9}{2}) = -2 - \frac{9}{2} = -\frac{13}{2}\).
04

Calculate the sum of the first 10 terms

Now substitute \(a_{10}\) into the sum formula: \(S_{10} = \frac{10}{2} \cdot (a + a_{10}) = 5 \cdot \left(-2 + \left(-\frac{13}{2}\right)\right) = 5 \cdot \left(-2 - \frac{13}{2}\right) = 5 \cdot \left(-\frac{4}{2} - \frac{13}{2}\right) = 5 \cdot \left(-\frac{17}{2}\right) = -\frac{85}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Arithmetic Series
When we talk about the sum of an arithmetic series, we're discussing the total of all terms up to a certain point in an arithmetic sequence. An arithmetic sequence is simply a list of numbers with a common difference between successive values.
To find the sum of the first \( n \) terms, we use the formula:
  • \( S_n = \frac{n}{2} \cdot (a + a_n) \)
In this formula:
  • \( S_n \) is the sum of the first \( n \) terms.
  • \( a \) is the first term.
  • \( a_n \) is the last term up to where we want to find the sum.
  • \( n \) is the number of terms.
First, you need to find the last term in the sequence (in our case, the 10th term), then plug it into the formula to get your answer. This nifty method saves time instead of adding each term individually.
Common Difference
The common difference in an arithmetic sequence is the fixed amount added or subtracted from one term to the next. It's what makes the sequence 'arithmetic'.
You can find this by subtracting any term from its preceding term. Using our example:
  • First term \( a = -2 \)
  • Second term = \(-\frac{5}{2}\)
So, the common difference \( d \) is:
  • \( d = -\frac{5}{2} - (-2) = -\frac{1}{2} \)
This tells us that each number in the sequence decreases by \(-\frac{1}{2}\) as you go from one term to the next. Understanding the common difference helps you see the pattern in the sequence.
Nth Term Formula
The nth term formula enables you to find any term in the sequence without listing all previous terms. This is super handy, especially for large sequences.
The formula is given by:
  • \( a_n = a + (n-1) \cdot d \)
Where:
  • \( a_n \) is the term you're looking for.
  • \( a \) is the first term.
  • \( n \) is the term's position.
  • \( d \) is the common difference.
For example, to find the 10th term, substitute the values:
  • \( a_n = -2 + (10-1)(-\frac{1}{2}) \)
Thus, \( a_{10} = -\frac{13}{2} \).
This formula gives you the power to quickly zero in on any specific term in the sequence.
Sequence Terms Calculation
Finding the terms in an arithmetic sequence involves using the common difference and the nth term formula we discussed. This allows you to list out as many terms as needed from the sequence without manual addition each time.
Here鈥檚 how you approach it:
  • Start with the first term, which is \(-2\) for our example.
  • Use the common difference, \(-\frac{1}{2}\), to get the next few terms.
For example:
  • Second term: \(-2 + (-\frac{1}{2}) = -\frac{5}{2}\)
  • Third term: \(-\frac{5}{2} + (-\frac{1}{2}) = -3\)
Continue this process using the nth term formula if you want to quickly jump to specific terms.
With practice, calculating sequence terms becomes second nature, and you鈥檒l be able to navigate sequences with ease.

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