Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Quiz for Lessons 12.1-12.3 - Page 817: 11


$a_n=\dfrac{3}{2}n-1$ $a_{15}=\dfrac{43}{2}$ $S_{15}=165$

Work Step by Step

The nth term is given by $a_n= a_1+(n-1) d$ ...(1) Here, we have Common Difference $d=\dfrac{3}{2}$ and first term $a_1=\dfrac{1}{2}$ Equation (1) gives: $a_n=\dfrac{1}{2}+\dfrac{3}{2} \times (n-1)=\dfrac{3}{2}n-1$ Plugging in $n =15$, we have $a_{15}=[\dfrac{3}{2} \times 15]-1=\dfrac{43}{2}$ We know that $S_{n}=\dfrac{n(a_1+a_{15})}{2}$ Now, $S_{15}=\dfrac{15 (\dfrac{1}{2}+\dfrac{43}{2})}{2}=165$
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