Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Chapter Test - Page 840: 10


Geometric; Sum: $\dfrac{125}{3}\left(1-\left(\dfrac{2}{5}\right)^n\right)$

Work Step by Step

We are given the sequence: $25,10,4,\dfrac{8}{5},....$ Determine the ratio of consecutive terms: $\dfrac{a_2}{a_1}=\dfrac{10}{25}=\dfrac{2}{5}$ $\dfrac{a_3}{a_2}=\dfrac{4}{10}=\dfrac{2}{5}$ $\dfrac{a_4}{a_3}=\dfrac{\dfrac{8}{5}}{4}=\dfrac{2}{5}$ As the ratio of consecutive terms is constant, the sequence is geometric. Its elements are: $a_1=25$ $d=\dfrac{2}{5}$ Determine the sum of the first $n$ terms: $S_n=a_1\cdot\dfrac{1-r^n}{1-r}$ $S_n=25\cdot\dfrac{1-\left(\dfrac{2}{5}\right)^n}{1-\dfrac{2}{5}}=25\cdot \left(1-\left(\dfrac{2}{5}\right)^n\right)\cdot\dfrac{5}{3}$ $=\dfrac{125}{3}\left(1-\left(\dfrac{2}{5}\right)^n\right)$
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