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: 48

Answer

First term: $a=195$ $n^{th}$ term: $a_n=195-7(n-1)$

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 twelfth term is $118$ . This means that: $a_{12} = a + d(12-1) \\118 = a + d(11) \\118 = a +11d$ The eight term is $146$. This means that: $a_8=a+d(8-1) \\146=a + d(7) \\146 = a + 7d$ The given information about the sequence results to two equations: (equation 1) $118=a+11d$ (equation 2) $146=a+7d$ Subtract equation 2 to equation 1 to obtain: $\begin{array}{cccc} & &118 &= &a +11d \\&-&&&&& \\& &146 &= &a + 7d \\&&\text{_____}&\text{______}&\text{_________} \\&&-28 &= &4d\end{array}$ Divide both sides by $4$ to obtain: $\dfrac{-28}{4}=\dfrac{4d}{4} \\-7 = d$ Solve for $a$ by substituting $d=-7$ to Equation 2 to obtain: $\require{cancel} 146 = a + 7d \\146 = a + 7(-7) \\146 = a + (-49) \\146=a-49 \\146+49=a \\195=a$ Thus, the first term is $a=195$. 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 $a= 195$ and $d=-7$. Substituting these to the nth term formula gives: $a_n = 195 + (-7)(n-1) \\a_n=195-7(n-1)$
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