Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015
1st Edition
ISBN: 9781680331165
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
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Chapter 8.2, Problem 62E

a.

To determine

To find: The value of n.

a.

Expert Solution
Check Mark

Answer to Problem 62E

The value of n=17 .

Explanation of Solution

Given information:

  i=1n(3i+5)=544

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  d=an+1an; n1 .

The nth tem of an arithmetic sequence with first term a1 and common difference d is given by : an=a1+(n1)d .

The sum of the first n terms of arithmetic series is

  Sn=n(a1+an2).

Calculation:

Consider ,

  i=1n(3i+5)=544

Step1. Calculating the first and last terms

  ai=3i+5a1=3(1)+5   =3+5   =8an=3n+5

Step2. The sum is given.

  =Sn=544=n(a1+an2)=544=n(8+3n+52)=544=n(3n+132)=544=3n2+13n=2(544)=3n2+13n1088=0

  =n=b±b24ac2a=n=13±1694(3)(1088)2(3)=n=13±169+130566=n=13±132256=n=13±1156

  =n=13+1156      =1026      =17 or

  =n=131156      =1286

Therefore , n=17 .

Hence, the number of terms are 17.

b.

To determine

To find: The value of n.

b.

Expert Solution
Check Mark

Answer to Problem 62E

The value of n=23 .

Explanation of Solution

Given information:

  i=1n(4i1)=1127

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  d=an+1an; n1 .

The nth tem of an arithmetic sequence with first term a1 and common difference d is given by : an=a1+(n1)d .

The sum of the first n terms of arithmetic series is

  Sn=n(a1+an2).

Calculation:

Consider ,

  i=1n(4i1)=1127

Step1. Calculating the first and last terms

  ai=4i1a1=4(1)1   =41   =5an=4n1

Step2. The sum is given .

  =n=b±b24ac2a=n=(6)±364(4)(2254)2(4)=n=6±36+36064(8)=n=6±36100(8)=n=6±190(8)

  =n=6+190(8)      =1968 or

  =n=6190(8)      =1848      =1848      =23

Therefore , n=23 .

Hence, the number of terms are 23.

c.

To determine

To find: The value of n.

c.

Expert Solution
Check Mark

Answer to Problem 62E

The value of n=9 .

Explanation of Solution

Given information:

  i=5n(7+12i)=455

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  d=an+1an; n1 .

The nth tem of an arithmetic sequence with first term a1 and common difference d is given by : an=a1+(n1)d .

The sum of the first n terms of arithmetic series is

  Sn=n(a1+an2).

Calculation:

Consider ,

  i=5n(7+12i)=455

The sum of the first 4 terms of arithmetic series is-

  i=14(7+12i)

Step1. The first and last terms

  ai=7+12ia1=7+12(1)   =7+12   =19a4=7+12(4)   =7+48    = 55

Step2. The sum is −

  Sn=n(a1+an2)S4=4(19+552)    =4(742)     =148

The sum of the first 4 terms is 148.

Now,

Sum of first n terms − Sum of first 4 terms = i=5n(7+12i)

Sum of first n terms − 148 = 455

Sum of first n terms = 455+148=603.

Now, consider

  i=1n(7+12i)=603

Step1. Calculating the first and last terms

  ai=7+12ia1=7+12(1)   =7+12   =19an=7+12n

Step2. The sum is given .

Now solving the equation we get,

  =n=b±b24ac2a=n=(26)±6764(12)(1,206)2(12)=n=26±676+57,88824=n=26±58856424=n=26±24224

  =n=26+24224      =21624      =9 or

  =n=2624224      =26824

Therefore , n=9 .

Hence, the number of terms are 9.

d.

To determine

To find: The value of n.

d.

Expert Solution
Check Mark

Answer to Problem 62E

The value of n=15 .

Explanation of Solution

Given information:

  i=3n(34i)=507

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d.

Therefore,

  d=an+1an; n1 .

The nth tem of an arithmetic sequence with first term a1 and common difference d is given by : an=a1+(n1)d .

The sum of the first n terms of arithmetic series is

  Sn=n(a1+an2).

Calculation:

Consider ,

  i=3n(34i)=507

The sum of the first 2 terms of arithmetic series is-

  i=12(34i)

Step1. The first and last terms

  ai=34ia1=34(1)   =34   =7a2=34(2)   =38    =11

Step2. The sum is −

  Sn=n(a1+an2)S2=2(7112)    =2(182)     =18

The sum of the first 2 terms is 18 .

Now,

Sum of first n terms − Sum of first 2 terms = i=12(34i)

Sum of first n terms − (18) = 507

Sum of first n terms = 50718=525.

Now, consider

  i=1n(34i)=525

Step1. Calculating the first and last terms

  ai=34ia1=34(1)   =34   =7an=34n

Step2. The sum is given .

Now solving the equation we get,

  =n=b±b24ac2a=n=(10)±1004(4)(1050)2(4)=n=10±100+168008=n=10±169008=n=10±1308

  =n=10+1308      =1408 or

  =n=101308      =1208      =15

Therefore , n=15 .

Hence, the number of terms are 15.

Chapter 8 Solutions

Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

Ch. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.1 - Prob. 27ECh. 8.1 - Prob. 28ECh. 8.1 - Prob. 29ECh. 8.1 - Prob. 30ECh. 8.1 - Prob. 31ECh. 8.1 - Prob. 32ECh. 8.1 - Prob. 33ECh. 8.1 - Prob. 34ECh. 8.1 - Prob. 35ECh. 8.1 - Prob. 36ECh. 8.1 - Prob. 37ECh. 8.1 - Prob. 38ECh. 8.1 - Prob. 39ECh. 8.1 - Prob. 40ECh. 8.1 - Prob. 41ECh. 8.1 - Prob. 42ECh. 8.1 - Prob. 43ECh. 8.1 - Prob. 44ECh. 8.1 - Prob. 45ECh. 8.1 - Prob. 46ECh. 8.1 - Prob. 47ECh. 8.1 - Prob. 48ECh. 8.1 - Prob. 49ECh. 8.1 - Prob. 50ECh. 8.1 - Prob. 51ECh. 8.1 - Prob. 52ECh. 8.1 - Prob. 53ECh. 8.1 - Prob. 54ECh. 8.1 - Prob. 55ECh. 8.1 - Prob. 56ECh. 8.1 - Prob. 57ECh. 8.1 - Prob. 58ECh. 8.1 - Prob. 59ECh. 8.1 - Prob. 60ECh. 8.1 - Prob. 61ECh. 8.1 - Prob. 62ECh. 8.1 - Prob. 63ECh. 8.1 - Prob. 64ECh. 8.2 - Prob. 1ECh. 8.2 - Prob. 2ECh. 8.2 - Prob. 3ECh. 8.2 - Prob. 4ECh. 8.2 - Prob. 5ECh. 8.2 - Prob. 6ECh. 8.2 - Prob. 7ECh. 8.2 - Prob. 8ECh. 8.2 - Prob. 9ECh. 8.2 - Prob. 10ECh. 8.2 - Prob. 11ECh. 8.2 - Prob. 12ECh. 8.2 - Prob. 13ECh. 8.2 - Prob. 14ECh. 8.2 - Prob. 15ECh. 8.2 - Prob. 16ECh. 8.2 - Prob. 17ECh. 8.2 - Prob. 18ECh. 8.2 - Prob. 19ECh. 8.2 - Prob. 20ECh. 8.2 - Prob. 21ECh. 8.2 - Prob. 22ECh. 8.2 - Prob. 23ECh. 8.2 - Prob. 24ECh. 8.2 - Prob. 25ECh. 8.2 - Prob. 26ECh. 8.2 - Prob. 27ECh. 8.2 - Prob. 28ECh. 8.2 - Prob. 29ECh. 8.2 - Prob. 30ECh. 8.2 - Prob. 31ECh. 8.2 - Prob. 32ECh. 8.2 - Prob. 33ECh. 8.2 - Prob. 34ECh. 8.2 - Prob. 35ECh. 8.2 - Prob. 36ECh. 8.2 - Prob. 37ECh. 8.2 - Prob. 38ECh. 8.2 - Prob. 39ECh. 8.2 - Prob. 40ECh. 8.2 - Prob. 41ECh. 8.2 - Prob. 42ECh. 8.2 - Prob. 43ECh. 8.2 - Prob. 44ECh. 8.2 - Prob. 45ECh. 8.2 - Prob. 46ECh. 8.2 - Prob. 47ECh. 8.2 - Prob. 48ECh. 8.2 - Prob. 49ECh. 8.2 - Prob. 50ECh. 8.2 - Prob. 51ECh. 8.2 - Prob. 52ECh. 8.2 - Prob. 53ECh. 8.2 - Prob. 54ECh. 8.2 - Prob. 55ECh. 8.2 - Prob. 56ECh. 8.2 - Prob. 57ECh. 8.2 - Prob. 58ECh. 8.2 - Prob. 59ECh. 8.2 - Prob. 60ECh. 8.2 - Prob. 61ECh. 8.2 - Prob. 62ECh. 8.2 - Prob. 63ECh. 8.2 - Prob. 64ECh. 8.2 - Prob. 65ECh. 8.2 - Prob. 66ECh. 8.2 - Prob. 67ECh. 8.2 - Prob. 68ECh. 8.2 - Prob. 69ECh. 8.2 - Prob. 70ECh. 8.2 - Prob. 71ECh. 8.2 - Prob. 72ECh. 8.2 - Prob. 73ECh. 8.3 - Prob. 1ECh. 8.3 - Prob. 2ECh. 8.3 - Prob. 3ECh. 8.3 - Prob. 4ECh. 8.3 - Prob. 5ECh. 8.3 - Prob. 6ECh. 8.3 - Prob. 7ECh. 8.3 - Prob. 8ECh. 8.3 - Prob. 9ECh. 8.3 - Prob. 10ECh. 8.3 - Prob. 11ECh. 8.3 - Prob. 12ECh. 8.3 - Prob. 13ECh. 8.3 - Prob. 14ECh. 8.3 - Prob. 15ECh. 8.3 - Prob. 16ECh. 8.3 - Prob. 17ECh. 8.3 - Prob. 18ECh. 8.3 - Prob. 19ECh. 8.3 - Prob. 20ECh. 8.3 - Prob. 21ECh. 8.3 - Prob. 22ECh. 8.3 - Prob. 23ECh. 8.3 - Prob. 24ECh. 8.3 - Prob. 25ECh. 8.3 - Prob. 26ECh. 8.3 - Prob. 27ECh. 8.3 - Prob. 28ECh. 8.3 - Prob. 29ECh. 8.3 - Prob. 30ECh. 8.3 - Prob. 31ECh. 8.3 - Prob. 32ECh. 8.3 - Prob. 33ECh. 8.3 - Prob. 34ECh. 8.3 - Prob. 35ECh. 8.3 - Prob. 36ECh. 8.3 - Prob. 37ECh. 8.3 - Prob. 38ECh. 8.3 - Prob. 39ECh. 8.3 - Prob. 40ECh. 8.3 - Prob. 41ECh. 8.3 - Prob. 42ECh. 8.3 - Prob. 43ECh. 8.3 - Prob. 44ECh. 8.3 - Prob. 45ECh. 8.3 - Prob. 46ECh. 8.3 - Prob. 47ECh. 8.3 - Prob. 48ECh. 8.3 - Prob. 49ECh. 8.3 - Prob. 50ECh. 8.3 - Prob. 51ECh. 8.3 - Prob. 52ECh. 8.3 - Prob. 53ECh. 8.3 - Prob. 54ECh. 8.3 - Prob. 55ECh. 8.3 - Prob. 56ECh. 8.3 - Prob. 57ECh. 8.3 - Prob. 58ECh. 8.3 - Prob. 59ECh. 8.3 - Prob. 60ECh. 8.3 - Prob. 61ECh. 8.3 - Prob. 62ECh. 8.3 - Prob. 63ECh. 8.3 - Prob. 64ECh. 8.3 - Prob. 65ECh. 8.3 - Prob. 66ECh. 8.3 - Prob. 67ECh. 8.3 - Prob. 68ECh. 8.3 - Prob. 69ECh. 8.3 - Prob. 70ECh. 8.4 - Prob. 1ECh. 8.4 - Prob. 2ECh. 8.4 - Prob. 3ECh. 8.4 - Prob. 4ECh. 8.4 - Prob. 5ECh. 8.4 - Prob. 6ECh. 8.4 - Prob. 7ECh. 8.4 - Prob. 8ECh. 8.4 - Prob. 9ECh. 8.4 - Prob. 10ECh. 8.4 - Prob. 11ECh. 8.4 - Prob. 12ECh. 8.4 - Prob. 13ECh. 8.4 - Prob. 14ECh. 8.4 - Prob. 15ECh. 8.4 - Prob. 16ECh. 8.4 - Prob. 17ECh. 8.4 - Prob. 18ECh. 8.4 - Prob. 19ECh. 8.4 - Prob. 20ECh. 8.4 - Prob. 21ECh. 8.4 - Prob. 22ECh. 8.4 - Prob. 23ECh. 8.4 - Prob. 24ECh. 8.4 - Prob. 25ECh. 8.4 - Prob. 26ECh. 8.4 - Prob. 27ECh. 8.4 - Prob. 28ECh. 8.4 - Prob. 29ECh. 8.4 - Prob. 30ECh. 8.4 - Prob. 31ECh. 8.4 - Prob. 32ECh. 8.4 - Prob. 33ECh. 8.4 - Prob. 34ECh. 8.4 - Prob. 35ECh. 8.4 - Prob. 36ECh. 8.5 - Prob. 1ECh. 8.5 - Prob. 2ECh. 8.5 - Prob. 3ECh. 8.5 - Prob. 4ECh. 8.5 - Prob. 5ECh. 8.5 - Prob. 6ECh. 8.5 - Prob. 7ECh. 8.5 - Prob. 8ECh. 8.5 - Prob. 9ECh. 8.5 - Prob. 10ECh. 8.5 - Prob. 11ECh. 8.5 - Prob. 12ECh. 8.5 - Prob. 13ECh. 8.5 - Prob. 14ECh. 8.5 - Prob. 15ECh. 8.5 - Prob. 16ECh. 8.5 - Prob. 17ECh. 8.5 - Prob. 18ECh. 8.5 - Prob. 19ECh. 8.5 - Prob. 20ECh. 8.5 - Prob. 21ECh. 8.5 - Prob. 22ECh. 8.5 - Prob. 23ECh. 8.5 - Prob. 24ECh. 8.5 - Prob. 25ECh. 8.5 - Prob. 26ECh. 8.5 - Prob. 27ECh. 8.5 - Prob. 28ECh. 8.5 - Prob. 29ECh. 8.5 - Prob. 30ECh. 8.5 - Prob. 31ECh. 8.5 - Prob. 32ECh. 8.5 - Prob. 33ECh. 8.5 - Prob. 34ECh. 8.5 - Prob. 35ECh. 8.5 - Prob. 36ECh. 8.5 - Prob. 37ECh. 8.5 - Prob. 38ECh. 8.5 - Prob. 39ECh. 8.5 - Prob. 40ECh. 8.5 - Prob. 41ECh. 8.5 - Prob. 42ECh. 8.5 - Prob. 43ECh. 8.5 - Prob. 44ECh. 8.5 - Prob. 45ECh. 8.5 - Prob. 46ECh. 8.5 - Prob. 47ECh. 8.5 - Prob. 48ECh. 8.5 - Prob. 49ECh. 8.5 - Prob. 50ECh. 8.5 - Prob. 51ECh. 8.5 - Prob. 52ECh. 8.5 - Prob. 53ECh. 8.5 - Prob. 54ECh. 8.5 - Prob. 55ECh. 8.5 - Prob. 56ECh. 8.5 - Prob. 57ECh. 8.5 - Prob. 58ECh. 8.5 - Prob. 59ECh. 8.5 - Prob. 60ECh. 8.5 - Prob. 61ECh. 8.5 - Prob. 62ECh. 8.5 - Prob. 63ECh. 8.5 - Prob. 64ECh. 8.5 - Prob. 65ECh. 8.5 - Prob. 66ECh. 8.5 - Prob. 67ECh. 8.5 - Prob. 68ECh. 8.5 - Prob. 69ECh. 8.5 - Prob. 70ECh. 8.5 - Prob. 71ECh. 8.5 - Prob. 72ECh. 8.5 - Prob. 73ECh. 8.5 - Prob. 74ECh. 8.5 - Prob. 75ECh. 8.5 - Prob. 76ECh. 8.5 - Prob. 77ECh. 8 - Prob. 1QCh. 8 - Prob. 2QCh. 8 - Prob. 3QCh. 8 - Prob. 4QCh. 8 - Prob. 5QCh. 8 - Prob. 6QCh. 8 - Prob. 7QCh. 8 - Prob. 8QCh. 8 - Prob. 9QCh. 8 - Prob. 10QCh. 8 - Prob. 11QCh. 8 - Prob. 12QCh. 8 - Prob. 13QCh. 8 - Prob. 14QCh. 8 - Prob. 15QCh. 8 - Prob. 16QCh. 8 - Prob. 17QCh. 8 - Prob. 18QCh. 8 - Prob. 19QCh. 8 - Prob. 1CRCh. 8 - Prob. 2CRCh. 8 - Prob. 3CRCh. 8 - Prob. 4CRCh. 8 - Prob. 5CRCh. 8 - Prob. 6CRCh. 8 - Prob. 7CRCh. 8 - Prob. 8CRCh. 8 - Prob. 9CRCh. 8 - Prob. 10CRCh. 8 - Prob. 11CRCh. 8 - Prob. 12CRCh. 8 - Prob. 13CRCh. 8 - Prob. 14CRCh. 8 - Prob. 15CRCh. 8 - Prob. 16CRCh. 8 - Prob. 17CRCh. 8 - Prob. 18CRCh. 8 - Prob. 19CRCh. 8 - Prob. 20CRCh. 8 - Prob. 21CRCh. 8 - Prob. 22CRCh. 8 - Prob. 23CRCh. 8 - Prob. 24CRCh. 8 - Prob. 25CRCh. 8 - Prob. 26CRCh. 8 - Prob. 27CRCh. 8 - Prob. 28CRCh. 8 - Prob. 29CRCh. 8 - Prob. 30CRCh. 8 - Prob. 31CRCh. 8 - Prob. 32CRCh. 8 - Prob. 33CRCh. 8 - Prob. 1CTCh. 8 - Prob. 2CTCh. 8 - Prob. 3CTCh. 8 - Prob. 4CTCh. 8 - Prob. 5CTCh. 8 - Prob. 6CTCh. 8 - Prob. 7CTCh. 8 - Prob. 8CTCh. 8 - Prob. 9CTCh. 8 - Prob. 10CTCh. 8 - Prob. 11CTCh. 8 - Prob. 12CTCh. 8 - Prob. 13CTCh. 8 - Prob. 14CTCh. 8 - Prob. 15CTCh. 8 - Prob. 1CACh. 8 - Prob. 2CACh. 8 - Prob. 3CACh. 8 - Prob. 4CACh. 8 - Prob. 5CACh. 8 - Prob. 6CACh. 8 - Prob. 7CA
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