Question

What is the 32nd term of the arithmetic sequence where a1= 4 and a6= 24?

1) 128
2) 132
3) 136
4) 140

Answer 1

i won't answer

Answer 2

marc this is wrong

Answer 3

angela don't answer

Answer 4

Marc...here are the formulas...please USE them
Arithmetic prog: a2-a1=d
an=an-1+d
an=a1+(n-1)*d
Sn=(a1+an)*n/2

Answer 5

why?

Answer 6

I wasn't...i got tired...I'm just giving the formulas

Answer 7

thank you

Answer 8

I have no clue how to use that formula though

Answer 9

Geometric prog:a2/a1=r
an=an-1*r
an=a1*r^(n-1)
Sn=a1*(r^n -1)/(r-1)

Answer 10

128

Answer 11

all u have to do is replace...(and take a look at ur math book)

Answer 12

Can you show me step by step with this problem what you mean by replace so i can see it better

Answer 13

u have a1 and a6 and u have to find a32,a32 from the formula would be a32=a1+(n-1)*d u don't have d so u have to find it from wht ur given
a6 from the formula is a6=a1+(n-1)*d
24=4+5d
5d=20
d=4(I'm supposing u know how to solve this)
a32 from the formula would be a32=a1+(n-1)*d=4+(32-1)*4=4+31*4=128

Answer 14

a1=a
Therefore a2=a+d
and a3=a+2d
a4=a+3d...
therefore a6=a+5d

Answer 15

Find a and d,
Then find a32=a+31d

Answer 16

a1=4 therefore a = 4
a6=a+5d therefore a+5d=24
but a=4 so 4+5d=24
therefore d=4
so to find a32,
a32=a+31d where a=4 and d=4
so a32= 4 + 31X4
which equals whatever

Answer 17

which equals 128 :)