Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - 12.2 Analyze Arithmetic Sequences and Series - 12.2 Exercises - Skill Practice - Page 807: 53

Answer

False

Work Step by Step

Let $a_n$ be the initial arithmetic series, $d_1$ its common difference, $b_n$ the second arithmetic series, $d_2$ its common difference and $n$ the number of terms in the sum. We are given: $$\begin{align*} d_2&=2d_1\\ b_1&=a_1. \end{align*}$$ The sum of each series is: $$\begin{align*} S_{n_1}&=\dfrac{n}{2}(a_1+a_n)\\ &=\dfrac{n}{2}(a_1+a_1+(n-1)d_1)\\ &=\dfrac{n}{2}(2a_1+(n-1)d_1). \end{align*}$$ $$\begin{align*} S_{n_2}&=\dfrac{n}{2}(b_1+b_n)\\ &=\dfrac{n}{2}(a_1+a_1+(n-1)d_2)\\ &=\dfrac{n}{2}(2a_1+2(n-1)d_1). \end{align*}$$ Calculate $S_2-2S_1$: $$\begin{align*} S_2-2S_1&=\dfrac{n}{2}(2a_1+2(n-1)d_1)-2\cdot \dfrac{n}{2}(2a_1+(n-1)d_1)\\ &=-na_1\not=0. \end{align*}$$ Since $S_1-2S_2\not=0$ it means that the sum of the series is not doubled. The given statement is FALSE.
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