Intermediate Algebra 12.1

Problem solving
69-70
Question 1
Nth term in the sequence is given by n+1
The first term will be therefore (1+1) ,(2+1) ,(3+1), (4+1), (5+1)
=1, 2, 3,4,5, 6
Question 3
The nth term is given by (n+3)/n
The series will be given by (1+3)/1, (2+3)/2, (3+3)/3, (4+3)/4, (5+3)/5
= 4, 2.5, 2, 1.75, 1.6
Question 5
Nth term is given by 3^n
= 3^1, 3^2, 3^3, 3^4, 3^5
= 3, 9, 27, 81, 243
Question 7
Nth term is given by 1/n2
= 1/12, 1/22, 1/32, 1/42, 1/52
= 1, 1/4, 1/9, 1/16, 1/25
Question 9
Nth term is given by 5(-1)n-1
The series will be
5(-1)1-1, 5(-1)2-1, 5(-1)3-1, 5(-1)4-1, 5(-1)5-1
5, -5, 5, -5, 5
Question11
The nth term is given by n-1/n
1-1/1, 2-1/2, 3-1/3, 4-1/4, 5-1/5
= 0, 1.5, 2.6667, 3.75, 4.8
Question 13
The nth term is given by
-9n+2
5th term will be given by
-9(5)+2
=-45+2
=-43
Question 15
Nth term is given by (3n+7)/(2n-5)
14th term will be
(3*14+7)/(2*14-5)
=49/23
=2.1304
Question 17
The nth term is given by
(n+1)(2n+3)
8th term is given by
=(8+1)(2*8+3)
=9*19
=171
Question 19
The series is 4, 8, 12, 16…
The general term representing the series for finding the nth term is given by
4n
Question 21
The series is -8,-18,-24,-32…
General term for the series is (-1)8n
Question 23
The series is 1/3, 1/9, 1/27, 1/81,
The general term for the series will be given by
1/3n
Question 25
2/5, 3/6, 4/7, 5/8….
The general term will be given by
(n+1)/(n+4)
Question 27
Borrowed $1000 installments of $100
There a…