Solutions for Beginning & Intermediate Algebra, 8th Edition by Lial - 2025 Published (All Chapters included)

Beginning & Intermediate Algebra,
8th Edition by Margaret L. Lial




Complete Chapter Solutions Manual
are included (Ch 1 to 14)




** Immediate Download
** Swift Response
** All Chapters included

,Table of Contents are given below




R. Prealgebra Review

1. The Real Number System

2. Linear Equations and Inequalities in One Variable

3. Linear Equations in Two Variables

4. Exponents and Polynomials

5. Factoring and Applications

6. Rational Expressions and Applications

7. Linear Equations, Graphs, and Systems

8. Inequalities and Absolute Value

9. Relations and Functions

10. Roots, Radicals, and Root Functions

11. Quadratic Equations, Inequalities, and Functions

12. Inverse, Exponential, and Logarithmic Functions

13. Nonlinear Functions, Conic Sections, and Nonlinear Systems

14. Further Topics in Algebra

,Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch14-1)
14.1 Sequences and Series 1243


Chapter 14 3. Make a table as follows:
Further Topics in Algebra Month Interest Payment Unpaid balance

0 10,000

14.1 Sequences and Series 1 10,000(0.02)  200 1000 + 200  120010,000  1000  9000

2 9000(0.02)  180 1000 + 180  1180 9000  1000  8000
Classroom Examples, Now Try Exercises
3 8000(0.02)  160 1000 + 160  1160 8000  1000  7000
1. an  2n  1 4 7000(0.02)  140 1000 + 140  1140 7000  1000  6000
To find a1 , the first term, replace n with 1.
The payments are $1200, $1180, $1160, and
a1  2 1  1  3 $1140; the unpaid balance is $6000.
To find a2 , the second term, replace n with 2. N3. Make a table as follows:
a2  2  2   1  5 Month Interest Payment Unpaid balance
To find a3 , the third term, replace n with 3. 0 8,000
a3  2  3  1  7 1 8000(0.02)  160 400 + 160  560 8000  400  7600

To find a4 , the fourth term, replace n with 4. 2 7600(0.02)  152 400 + 152  552 7600  400  7200

a4  2  4   1  9 3 7200(0.02)  144 400 + 144  544 7200  400  6800

To find a5 , the fifth term, replace n with 5. 4 6800(0.02)  136 400 + 136  536 6800  400  6400

a5  2  5   1  11 The payments are $560, $552, $544, and $536;
the unpaid balance is $6400.
N1. an  5  3n 5
To find a1 , the first term, replace n with 1. 4. (a)   2i  4 
a1  5  3 1  2
i 1

  2 1  4    2  2  4 
To find a2 , the second term, replace n with 2.
  2  3  4   2  4  4
a2  5  3  2   1
  2  5  4
To find a3 , the third term, replace n with 3.
 2  0  2  4  6
a3  5  3  3  4
 10
To find a4 , the fourth term, replace n with 4.
6
a4  5  3  4   7 (b)   i  1 i  2 
i 2
To find a5 , the fifth term, replace n with 5.
  2  1 2  2    3  1 3  2 
a5  5  3  5   10
  4  1 4  2    5  1 5  2 

2.
1
,
1 1 1
, , , ... can be written as   6  1 6  2 
2 4 8 16
 3  0   4 1  5  2   6  3  7  4 
1 1 1 1 1
1
, 2 , 3 , 4 , ... , so an  n , or  0  4  10  18  28
2 2 2 2 2
n  60
1
an    . 4
2   2i 
2
(c)
i 1

  2 1   2  2    2  3   2  4 
2 2 2 2
N2. 3, 9,  27, 81, ... can be written as
 31 ,  32 ,  33 ,  34 , ... , so an   3  .
n
 2 2  4 2  6 2  82
 4  16  36  64
 120

, 1244 Chapter 14 Further Topics in Algebra

5 n 5
N4.  i  4 2
 xi  xi
i 1 i 1
N6. x   i 1
 1  4    2  4    3
2 2 2
4  n 5
 439, 714  167,367  114,123 
  4  4  5  4
2 2 
 96,535  91, 477

 
 1  4    4  4    9  4  5
 16  4    25  4  909, 216

5
 3  0  5  12  21
 181,843.2
 35
To the nearest whole number, the average
5. (a) 7  12  17  22 number of Quarter Horses registered per state
The terms increase by 5, so 5i is part of the in the top five states is 181,843.
general term. If i  1, 5i  5, but since the
first term is 7, we must add 2 to 5i to get 7. Exercises
There are four terms, so 1. The domain of an infinite sequence is the set of
4
all positive integers (natural numbers).
7  12  17  22    5i  2  .
i 1
2. In the sequence 3, 6, 9, 12, the term a3  9.

(b) 3  12  27  48  75 3. If an  2n, then a40  80.
 3(1  4  9  16  25)
4. If an   1 , then a115  1.
n

 3 12  22  32  42  52 
3

  i  2  is 12.
5
  3i 2 5. The value of the sum
i 1 i 1


N5. (a) 3  5  7  9  11 6. The arithmetic mean of 4,  2, 0, 2, and 4 is 0.
The terms increase by 2, so 2i is part of the
7. an  n  1
general term. If i  1, 2i  2, but since the
first term is 3, we must add 1 to 2i to get 3. a1  1  1  2 Replace n with 1.
There are five terms, so a2  2  1  3 Replace n with 2.
5
a3  3  1  4 Replace n with 3.
3  5  7  9  11    2i  1
i 1 a4  4  1  5 Replace n with 4.
(b) 1  4  9  16  25 a5  5  1  6 Replace n with 5.
 12  2 2  32  4 2  52 Answer: 2, 3, 4, 5, 6
5
8. an  n  4
  i 2
i 1 a1  1  4  5 Replace n with 1.
n 5 a2  2  4  6 Replace n with 2.
 xi  xi a3  3  4  7 Replace n with 3.
i 1 i 1
6. x   a4  4  4  8 Replace n with 4.
n 5
5671  5409  5183  5005  4844 a5  5  4  9 Replace n with 5.

5 Answer: 5, 6, 7, 8, 9
26,112

5
 5222.4
To the nearest unit, the average number of
FDIC-insured institutions for the 5-yr period
2017–2021 is 5222.