number patterns?

Question

what is the rule for this sequence?:
2, 4, 10, 28, 82...
Then what are the two numbers after?
And also what is the 10th term?

Answer 1

multiply by 3 then subtract 2

the next two numbers are 244 and 730

the 10th term (number) is 19684

Answer 2

4 -2 = 2
10 - 4 = 6 (2 X 3)
28 - 10 = 18 (6 X 3)
82 - 28 = 54 (18 X 3)
82 + (54 X 3) = 82 + 162 = 244
244 + (162 X 3) = 244 + 486 = 730

I'll let you figure out the tenth term from there.

Answer 3

multiply by 3, subtract 2

2 x 3 = 6 -2 =4
4 x 3 = 12 -2 = 10

6th term: 82 x 3 = 246 -2 = 244
7th term :244 x 3 = 632 -2 = 630

10th term : 16984 (check my arithmetic, though)

Answer 4

I can't remember the proper way to write it, but the next number is 3x-2
2*3=6 6-2=4
4*3=12 12-2=10
10*3=30 30-2=28
and so on,
so the next numbers are 244, 730, 2188, 6562, 19684

Answer 5

2 + 2 = 4
4 + 2 * 3 = 10
10 + 2 * 3 * 3 = 28
28 + 2 * 3 * 3 * 3 = 82
82 + 2 * 3 * 3 * 3 * 3 = 244
244 + 2 * 3 * 3 * 3 * 3 * 3 = 730
Rule can't be based on the number that came before it. Thus, it can be seen that the rule has somthing to do with 3 to the power of something.
When examined, the equation is:
K = (3 ^ (n-1)) + 1
K = [3 ^ (10 - 1)] + 1
K = 19684

Answer 6

The rule is 3^n+1 for n>=1. The next two numbers are 244 and 730. The 10th term is 19684.

http://www.research.att.com/~njas/sequences/?q=2%2C+4%2C+10%2C+28%2C+82&language=english&go=Search

For other sequences, see http://www.research.att.com/~njas/sequences/Seis.html

Answer 7

2, 4, 10, 28, 82...

f(1)=2=(3^(1-1))+1
f(2)=4=(3^(2-1))+1
f(3)=10=(3^(3-1))+1
f(4)=28=(3^(4-1))+1
f(5)=82=(3^(5-1))+1
f(n)=(3^(n-1))+1

so f(6)=(3^(6-1))+1=244
f(7)=(3^(7-1))+1=730

f(10)=(3^(10-1))+1
=19683+1=19684