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Chapter 14: Problem 60

Use a system of two equations in two variables, \(a_{1}\) and \(d,\) to solve. Write a formula for the general term (the \(n\) th term) of the arithmetic sequence whose third term, \(a_{3},\) is 7 and whose eighth term, \(a_{\mathrm{g}},\) is 17

Short Answer

Expert verified
The formula for the nth term of the arithmetic sequence is \(a_n = 2n + 1.\

Step by step solution

01

Formulating Equations

In an arithmetic sequence, the nth term could be expressed with the formula \(a_n = a_1 + (n-1) \cdot d \). From the problem, we know that \(a_3 = 7\) and \(a_8 = 17.\) We can substitute these values in the nth term formula to get two equations: \n1. \(7 = a_1 + 2d \) \n2. \(17 = a_1 + 7d\)
02

Solving the system of equations

We can solve the system of equations above for \(a_1\) and \(d\), the first term and the common difference of the sequence. Subtract the first equation from the second equation to isolate \(d\): \n\(17 - 7 = (a_1 + 7d) - (a_1 + 2d) \) \nDividing the result by 5, we find that \(d = 2.\) Substitute \(d = 2\) into the first equation to solve for \(a_1\): \n\(7 = a_1 + 2(2) => a_1 = 7 - 4 = 3\)
03

Writing the general formula

Knowing the first term, \(a_1 = 3,\) and the common difference, \(d = 2,\) we can formulate the general term of the arithmetic sequence: \(a_n = a_1 + (n-1) \cdot d = 3 + (n-1) \cdot 2 = 2n +1.\

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Key Concepts

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

System of Equations
Understanding a system of equations is fundamental in tackling problems involving arithmetic sequences and beyond. In the context of sequences, specifically arithmetic ones, a system of equations allows us to solve for multiple unknowns that are related through several conditions. In the given exercise, we have two unknowns: the first term of the sequence, denoted as a1, and the common difference, d.

When forming a system of equations, we use known values to create simultaneous equations. Here, we know the third term a3 and the eighth term a8. By substituting these into the nth term formula, which relates every term in the sequence to a1 and d, we can create two algebraic equations that, when solved together, yield the values of our unknowns. Solving systems typically involves methods like substitution, elimination, or graphing鈥攂ut in this case, simple subtraction is sufficient to find the common difference.
Nth Term Formula
The nth term formula of an arithmetic sequence is the blueprint for finding any term in the sequence. It is generally denoted as an = a1 + (n - 1)d, where a1 is the first term, d is the common difference, and n signals the position of the term within the sequence.

The power of the nth term formula lies in its ability to provide a direct computation for any term, given a1 and d. Once we identify these two key components, no matter the size of n, the formula will reliably predict the value of that term. This is immensely useful not just for small sequences, but also for understanding long-term behaviors of sequences in mathematics and applied sciences.
Common Difference
The common difference is possibly the most definitive feature of an arithmetic sequence. It's the consistent interval between consecutive terms, represented by d in our expressions. Identifying the common difference gives us insight into the pattern of the sequence and allows us to predict or calculate terms that may not be immediately obvious.

The common difference is what makes an arithmetic sequence 'arithmetic'. It remains constant, and when added to any term in the sequence, results in the next term. In the exercise, finding the common difference was a matter of simple arithmetic: we subtracted the known terms and divided by the difference in their positions within the sequence. Knowing the common difference, along with the first term of the sequence, unlocks the potential to find any term in the sequence using the nth term formula.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The sum of the geometric series \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{512}\) can only be estimated without knowing precisely what terms occur between \(\frac{1}{8}\) and \(\frac{1}{512}\).

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. There's no end to the number of geometric sequences that I can generate whose first term is 5 if I pick nonzero numbers \(r\) and multiply 5 by each value of \(r\) repeatedly.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If the \(n\) th term of a geometric sequence is \(a_{n}=3(0.5)^{n-1}\) the common ratio is \(\frac{1}{2}\).

Will help you prepare for the material covered in the next section. Consider the sequence \(8,3,-2,-7,-12, \ldots .\) Find \(a_{2}-a_{1}, a_{3}-a_{2}, a_{4}-a_{3},\) and \(a_{5}-a_{4} .\) What do you observe?

Each exercise involves observing a pattern in the expanded form of the binomial expression \((a+b)^{n}\).$$\begin{array}{l}(a+b)^{1}=a+b \\\\(a+b)^{2}=a^{2}+2 a b+b^{2} \\\\(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\\\(a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\\\(a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5}\end{array}$$ Describe the pattern for the sum of the exponents on the variables in each term.
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