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 are ten installments
The first month will be 100+ (0.01*1000)
The second month they will pay 100+ (0.01*900)
=109, the third month 100+ (0.01*800) =108 the fourth month will be 100+(0.01*700)=107
The sequence is 110, 109,108, 107, 106, 105, 104, 103, 102, 101The first six months they will have paid as follows
110, 109,108, 107, 106, 105,
The remaining debt balance will be $400. However it will be paid with some interest.
Question 29
The end of first year, the value will be 4/5 of 20000
=16000
The second year=4/5*16000=12800
The third year 0.

8or 4/5 *12800
=10240
The fourth year the value will be 4/5*10240=8192
The fifth year the value will be 4/5*8192
=6554
Question 31
∑(i+3) from 1 to 5
=(1+3) +(2+3) +(3+3)+ (4+3) +(5+3)
=4+5+6+7+8
=30
Question 33
∑(i2+2) from 1 to 3
=(12+2)+(22+2)+(32+2)
=3+6+11
=20
Question 35
∑(-1)-I from 1 to 6
(-1)1+(-1)2+ (-1)3 + (-1)4 +(-1)5+(-1)6
=-1+1+-1+1+-1+1
=0
Question 37
∑(i-3)(i+2) from 3 to 7
=[(3-3)(3+2)] +[(4-3)(4+2)] +[(5-3)(5+2)]+[(6-3)(6+2)]+[(7-3)(7+2)]
=0+6+14+24+36
=80
Question 39
3+4+5+6+7
∑(i+2) from i=1 to i =5
Question 41
-2+4-8+16-32
=∑(-2)i where i is from 1 to 5
Question 43
1+4+9+16
It will be given by
∑(i)2 where i starts from 1 to 4
Question 45
The difference between a sequence and a series is that a sequence is a pattern followed in in coming up with the terms. Terms are written in a certain pattern. While a series is the sum of the terms of the sequence.71
Question 3
The sequence is 1/3, 2/3, 3/3, 4/3,
The series is not geometric.
Question 5
The sequence is a geometrical
1, -3, 9, -27, 81….
The ratio is given by second term/ first term
=-3/1=-3
Question 7
1, -1/2, 1/4, -1/8, 1/16….
The sequence is geometrical with the r ratio being
(-1/2)/1 = -1/2
Question 9
Find the general term of the sequence
-5,-10,-20
The ratio is given by -10/-5=2
Therefore the general term is
(-5)*2i where I starts from 0 to 2
(-5)*2n-1
Question 11
-2, 2/3, -2/9
The common ratio is ( 2/3)/-2
=-1/3
The general term is (-2)*(-1/3)I where I starts from 0 to 2
(-2)*(-1/3)n-1
Question 13
The sequence is 10, -2, 2/5…..
The ratio is -2/10
=-1/5
The general term is 10*(-1/5)I where I starts from 0 to 2
Or 10*(-1/5)n-1
Question 15
Nth term in a geometrical sequence is given by arn where a is the first term and r is the common ration
in this case a=2 and r =5
the 10th term will be
2*510-1
=3906250
Question 17
½, 1/6/ 1/18 this is the sequence
The first term is ½ while the common ratio is 1/6*2/1
=1/3.
12 term will be given by1/2*(1/3)12-1
=0.000002823
Question 19
The third term is ½ the 7th term is 1/32 the 25th term will be
1/2=ar2
1/32=ar6
Finding the ratio of the two equations we get
R4=1/16
R=1/2
1/2=a(1/2)2
1/2=a*1/4
a=2
the 25th term = 2*(1/2)25-1
=1/8388608
Question 21
A=2 r=3
The first five terms of the sequence will be given by
a, ar, ar2, ar3, ar4
=2, 2*3, 2*32, 2*33, 2*34
=2, 6, 18, 54, 162
Question 23
A=5 r =-1/5
The first five terms of the sequence will be given by
a, ar, ar2, ar3, ar4
5, 5*(-1/5), 5(-1/5)2, 5(-1/5)3, 5(-1/5)4
=5, -1, 1/5, -1/25, 1/125
Question 25
Sn= a(1-rn)/(1-r)
First term is 1/3 and the common ratio is (1/9)/(1/3)
Ratio is 1/3
=1/3(1-(1/3)5)/(1-1/3)
0.498
Question 27
-4/3, -4/9, -4/27, -4/ 81, -4/ 243, -4/ 729
Ratio= (-4/9)*(-3/4) =1/3
A=-4/3
-4/3(1-(1/3)6)/(1-1/3)
=1.997
Question 29
∑4(2/5)I where I start from 1 to 7
4(2/5) +4(2/5)2+4(2/5)3+4(2/5)4+4(2/5)5+4(2/5)6+4(2/5)6+4(2/5)7
=-2.662
Question 31
∑-2(3/5)I where I starts from 1 to 10
The first few terms are given by
-6/5, -18/25…..
A= -6/5 ratio = -18/25*-5/6=3/5
S10= -6/5(1-(3/5)10)/1-3/5
=-6/5(0.994)
=-2.982
72 7,13,21,35,37Question 7
4!/0!4!
=(4*3*2*1)(4*3*2*1)
=1
Question 13
13C11
=13!/(11!(13-11)!
=13*12*/(4*3*2)
=13*12/24
=6.5
Question 21
(x/2-y)4
X4y0+x3y1+x2y2+ x1y3+x0+y4
Coefficients
are 1, 4, 6, 4,1
1X4y0+4x3y1+6x2y2+ 4x1y3+ 1×0+y4
Where y=-y x= x/2
X4/16-4[(x3/8)y]+6[(x2/4)y2)-4[(x/2)y3+y4
X4/16-[(x3/2)y]+3[(x2/2)y2)-2(xy3)+y4
Question 35
(x+y/2)8 seventh term.=X1Y6
The coefficient will be
8c7= 8
8 X1Y6
8xy6/64
=8xy6
Question 37
(k-1)9 the third term
Third term is X7y2
Coefficient 84
84 X7y2
84K7