## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Review Exercises - Page 1124: 32

#### Answer

$a_{1}=\dfrac{1}{2};a_2=\dfrac{1}{4};a_3=\dfrac{1}{8};a_4=\dfrac{1}{16};a_5=\dfrac{1}{32}$

#### Work Step by Step

The general formula to find the nth term of a Geometric sequence is given as: $a_{n}=a_1r^{n-1}$ Here, $a_n=\dfrac{1}{2}(\dfrac{1}{2})^{n-1}$ Plug $n=1,2,3,4,5$ $a_{1}=\dfrac{1}{2}(\dfrac{1}{2})^{1-1}=\dfrac{1}{2};a_2=\dfrac{1}{2}(\dfrac{1}{2})^{2-1}=\dfrac{1}{4};a_3=\dfrac{1}{2}(\dfrac{1}{2})^{3-1}=\dfrac{1}{8};a_4\dfrac{1}{2}(\dfrac{1}{2})^{4-1}=\dfrac{1}{16};a_5=\dfrac{1}{2}(\dfrac{1}{2})^{5-1}=\dfrac{1}{32}$ So, the first $5$ terms of the sequence are $a_{1}=\dfrac{1}{2};a_2=\dfrac{1}{4};a_3=\dfrac{1}{8};a_4=\dfrac{1}{16};a_5=\dfrac{1}{32}$

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