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: 56

Answer

$S_{15}=165$

Work Step by Step

Note that from $a_2$, you have to add the common difference thrice to get the value of $a_5$. This means that $a_5=a_2+3d$ With $a_2=8$ and $a_5=9.5$, substituting these values to the equation above gives: $a_5=a_2+3d \\9.5=8+3d \\9.5-8=8+3d-8 \\1.5=3d \\\dfrac{1.5}{3} = \dfrac{3d}{3} \\0.5=d$ With $d=0.5$ and $a_2=8$, the first term can be found by subtracting $d$ to $a_2$: $a = a_2 -d \\a=8-0.5 \\a=7.5$ RECALL: The sum of the first $n$ terms of an arithmetic sequence is given by the formula: $S_n=\dfrac{n}{2}[2a + (n-1)d]$ where $a$ = first term $d$ = common difference Using the formula above, the sum of the first 15 terms of the given arithmetic sequence is: $S_n = \dfrac{n}{2}[2a + (n-1)d] \\S_{15}=\dfrac{15}{2}[2(7.5)+(15-1)0.5] \\S_{15}=\dfrac{15}{2}[15+14(0.5)] \\S_{15}=\dfrac{15}{2}(15+7) \\S_{15}=\dfrac{15}{2}(22) \\S_{15}=165$
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