2. Isn't a1=-3, and d=+6 ?
3--I have no idea what the question is asking.
5.
Sum= sigma(k=3,8) of (40-3k)
(40-3*3)+(40-3*4) + ... as you wrote the terms.
It will break down to
40*(8-2) - 3(3+4+5+6+7+8)=240-3*33=240-99= 141 as upi determined.
=
1)Find the 20th term of the arithmetic sequence in which a1=3 and d=7
a.143
b.136
c.140
d.133
an=a1+(n-1)d
a20=3+(20-1)7
a20=3+(19)7
a20=3+133
answer=b
2)Write an equation for the nth term of the arithmetic sequence -3,3,9,15...
a.an=n+6
b.an=6n+9
c.an=6n-9
d.an=n-3
an=a1+(n-1)d
an=3+(n-1)-6
an=3-6n+6
answer=b
3)Find the two arithmetic means between 4 and 22
a.10,16
b.8,16
c.8,12
d.13,13
an=a1+(n-1)d
a6=4+(6-1)d
22=4+6d
answer=a
4)Simplify:Find Snfor the arithmetic series in which a1=3,d=1/2,and an=15
a.225
b.9
c.45
d.210
Sn=n/2(a1+an)
Sn=15/2(3+15)
answer=b
5)I don't know how to type this but its like this in my book.
Find; 8 is on top of the sigma notation and k=3 is at the bottom(40-3k) is on the right side
a.45
b.282
c.-90
d.141
40-3(3)=40-9=31
40-3(4)=40-12=28
40-3(5)=40-15=25
40-3(6)=40-18=22
40-3(7)=40-21=19
40-3(8)=40-24=16
answer=d
1 answer
1 answer