EBK PRECALCULUS W/LIMITS

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Concept explainers

Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Question

Chapter 9, Problem 5PS

To determine

State the common difference.

Expert Solution

Answer to Problem 5PS

$d$ .

Explanation of Solution

Calculation:

Consider the arithmetic sequence with first term $a$ and common difference $d$ .

$a,a+d,a+2d,a+3d,...$

The general term $nth$ of the sequence is given by

$an=a+(n−1)d$

Add a constant $C$ to each term of the sequence.

So, the resulting sequence will be

$a+C,a+d+C,a+2d+C,a+3d+C,...$

Then the general term of the sequencing will be

$an=a+(n−1)d+C$

Note that ,

$an−an−1=(a+(n−1)d+C)−(a+(n−2)d+C)$

$=d$

Difference between each two consecutive term is same,

Hence, the resulting sequence is also an arithmetic sequence and the coomon difference of arithmetic sequence is $d$ .

To determine

State the common difference.

Expert Solution

Answer to Problem 5PS

$Cd$ .

Explanation of Solution

Calculation:

Consider the arithmetic sequence with first term $a$ and common difference $d$ .

$a,a+d,a+2d,a+3d,...$

The general term $nth$ of the sequence is given by

$an=a+(n−1)d$

Multiply each term of the sequence by $C$ .

So, the resulting sequence will be

$aC,(a+d)C,(a+2d)C,(a+3d)C,...$

Then the general term of the sequencing will be

$an=(a+(n−1)d)C$

Note that ,

$an−an−1=(a+(n−1)d)C−(a+(n−2)d)C$

$=Cd$

Difference between each two consecutive term is same,

Hence, the resulting sequence is also an arithmetic sequence and the coomon difference of arithmetic sequence is $Cd$ .

To determine

Find if each operation results in an arithmetic sequence.

Expert Solution

Answer to Problem 5PS

The sequence is not an arithmetic sequence.

Explanation of Solution

Calculation:

Consider the arithmetic sequence with first term $a$ and common difference $d$ .

$a,a+d,a+2d,a+3d,...$

The general term $nth$ of the sequence is given by

$an=a+(n−1)d$

Square each term of the sequence.

So, the resulting sequence will be

$a2,(a+d)2,(a+2d)2,(a+3d)2,...$

Then the general term of the sequencing will be

$an=(a+(n−1)d)2$

Note that ,

$an−an−1=(a+(n−1)d)2−(a+(n−2)d)2=(a2 +(n−1)2d2+2a(n−1)d−(a2 +(n−2)2d2+2a(n−2)d))=(−3+2n)d2+2ad$

So, the difference depends upon the value $n$ .

Difference between each two consecutive terms is not same.

Hence, the resulting sequence is not arithmetic sequence.

Chapter 9, Problem 4PS

Chapter 9, Problem 6PS

Chapter 9 Solutions

EBK PRECALCULUS W/LIMITS

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Prob. 100RE Ch. 9 - Prob. 101RE Ch. 9 - Prob. 102RE Ch. 9 - Prob. 103RE Ch. 9 - Prob. 104RE Ch. 9 - Prob. 105RE Ch. 9 - Prob. 106RE Ch. 9 - Prob. 107RE Ch. 9 - Prob. 108RE Ch. 9 - Prob. 1CT Ch. 9 - Prob. 2CT Ch. 9 - Prob. 3CT Ch. 9 - Prob. 4CT Ch. 9 - Prob. 5CT Ch. 9 - Prob. 6CT Ch. 9 - Prob. 7CT Ch. 9 - Prob. 8CT Ch. 9 - Prob. 9CT Ch. 9 - Prob. 10CT Ch. 9 - Prob. 11CT Ch. 9 - Prob. 12CT Ch. 9 - Prob. 13CT Ch. 9 - Prob. 14CT Ch. 9 - Prob. 15CT Ch. 9 - Prob. 16CT Ch. 9 - Prob. 17CT Ch. 9 - Prob. 18CT Ch. 9 - Prob. 19CT Ch. 9 - Prob. 20CT Ch. 9 - Prob. 1CLT Ch. 9 - Prob. 2CLT Ch. 9 - Prob. 3CLT Ch. 9 - Prob. 4CLT Ch. 9 - Prob. 5CLT Ch. 9 - Prob. 6CLT Ch. 9 - Prob. 7CLT Ch. 9 - Prob. 8CLT Ch. 9 - Prob. 9CLT Ch. 9 - Prob. 10CLT Ch. 9 - Prob. 11CLT Ch. 9 - Prob. 12CLT Ch. 9 - Prob. 13CLT Ch. 9 - Prob. 14CLT Ch. 9 - Prob. 15CLT Ch. 9 - Prob. 16CLT Ch. 9 - Prob. 17CLT Ch. 9 - Prob. 18CLT Ch. 9 - Prob. 19CLT Ch. 9 - Prob. 20CLT Ch. 9 - Prob. 21CLT Ch. 9 - Prob. 22CLT Ch. 9 - Prob. 23CLT Ch. 9 - Prob. 24CLT Ch. 9 - Prob. 25CLT Ch. 9 - Prob. 26CLT Ch. 9 - Prob. 27CLT Ch. 9 - Prob. 28CLT Ch. 9 - Prob. 29CLT Ch. 9 - Prob. 30CLT Ch. 9 - Prob. 31CLT Ch. 9 - Prob. 32CLT Ch. 9 - Prob. 33CLT Ch. 9 - Prob. 34CLT Ch. 9 - Prob. 35CLT Ch. 9 - Prob. 36CLT Ch. 9 - Prob. 37CLT Ch. 9 - Prob. 38CLT Ch. 9 - Prob. 39CLT Ch. 9 - Prob. 40CLT Ch. 9 - Prob. 41CLT Ch. 9 - Prob. 1PS Ch. 9 - Prob. 2PS Ch. 9 - Prob. 3PS Ch. 9 - Prob. 4PS Ch. 9 - Prob. 5PS Ch. 9 - Prob. 6PS Ch. 9 - Prob. 7PS Ch. 9 - Prob. 8PS Ch. 9 - Prob. 9PS Ch. 9 - Prob. 10PS Ch. 9 - Prob. 11PS Ch. 9 - Prob. 12PS Ch. 9 - Prob. 13PS Ch. 9 - Prob. 14PS Ch. 9 - Prob. 15PS Ch. 9 - Prob. 16PS

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