College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.2 - Arithmetic Sequences - 8.2 Exercises: 47

Answer

$a=-\dfrac{5}{12}$ $a_n = -\dfrac{5}{12} + \dfrac{1}{4}(n-1)$

Work Step by Step

RECALL: The $n^{th}$ $a_n$ of an arithmetic sequence is given by the formula: $a_n = a + d(n-1)$ where $a$ = first term $d$ = common difference The fourteenth term is $\frac{2}{3}$ . This means that: $a_{14} = a + d(14-1) \\\frac{2}{3} = a + d(13) \\\frac{2}{3} = a +13d$ The ninth term is $\frac{1}{4}$. This means that: $a_9=a+d(9-1) \\\frac{1}{4}=a + d(8) \\\frac{1}{4} = a + 8d$ The given information about the sequence results to two equations: (equation 1) $\frac{2}{3}=a+13d$ (equation 2) $\frac{1}{4}=a+8d$ Subtract equation 2 to equation 1 to obtain: $ \begin{array}{cccc} & &\frac{2}{3} &= &a +13d \\&-&&&&& \\& &\frac{1}{4} &= &a + 8d \\&&\text{_____}&\text{______}&\text{_________} \\&&\frac{5}{12} &= &5d\end{array}$ Divide both sides by $5$ to obtain: $\dfrac{\frac{5}{12}}{5}=\dfrac{5d}{5} \\\frac{5}{12} \cdot \frac{1}{5} = d \\\frac{1}{12}=d$ Solve for $a$ by substituting $d=\frac{1}{12}$ to Equation 2 to obtain: $\require{cancel} \frac{1}{4} = a + 8d \\\frac{1}{4} = a + 8(\frac{1}{12}) \\\frac{1}{4} = a + \frac{8}{12} \\\frac{1}{4} = a + \frac{2\cancel{(4)}}{3\cancel{(4)}} \\\frac{1}{4} = a + \frac{2}{3} \\\frac{1}{4} - \frac{2}{3} = a \\\frac{3}{12} = \frac{8}{12} = a \\\frac{3-8}{12}=a \\\frac{-5}{12}=a$ Thus, the first term is $a=-\dfrac{5}{12}$. The $n^{th}$ term $a_n$ of an arithmetic sequence is given by the formula: $a_n=a+d(n-1)$ where $a$ = first term $d$ = common difference The given sequence has $a= -\dfrac{5}{12}$ and $d=\dfrac{1}{12}$. Substituting these to the nth term formula gives: $a_n = -\dfrac{5}{12} + \dfrac{1}{4}(n-1)$
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