PDA

View Full Version : Geometric sequence

boditree
12-06-2009, 01:48 AM
Is this a geometric sequence? I cannot find a common ratio.

10, 16, 23, 31, 40, 50, 61,...

wjm11
12-06-2009, 03:18 AM
Is this a geometric sequence? I cannot find a common ratio.

10, 16, 23, 31, 40, 50, 61,...

You have correctly answered your own question; it has no common ratio, so it is not a geometric sequence.

PS If you examine the differences between terms, you'll see a pattern: add 6,7,8,9,10,11...

boditree
12-06-2009, 12:03 PM
Thanks wjm11, I saw the pattern. The trick is putting it in a formula to predict to the n th term.

chrisr
12-06-2009, 01:25 PM
T[sub:1pa4bbel]1[/sub:1pa4bbel]=10
T[sub:1pa4bbel]2[/sub:1pa4bbel]=10+6=10+1(6)
T[sub:1pa4bbel]3[/sub:1pa4bbel]=10+6+7=10+6+6+1=10+2(6)+1
T[sub:1pa4bbel]4[/sub:1pa4bbel]=10+6+7+8=10+6+6+6+1+2=10+3(6)+(1+2) .....

from there you can begin to formulate

Subhotosh Khan
12-06-2009, 02:25 PM
Thanks wjm11, I saw the pattern. The trick is putting it in a formula to predict to the n th term.

With that pattern - a second-order function is required.

a_n = \frac{1}{2}\cdot (n^2 + 9n + 10)

chrisr
12-06-2009, 03:53 PM
That's it.

T[sub:5boe7kh7]n[/sub:5boe7kh7]=10+(n-1)6+the sum from i=1 to n-2 of i.
This sum is very simple, it's arithmetic.
The sum of an arithmetic series is (average value)(number of terms).
The average is (first+last)/2.
In this case, the average is (1+n-2)/2 =(n-1)/2 and there are n-2 terms.

Hence T[sub:5boe7kh7]n[/sub:5boe7kh7]=10 + (n-1)6 + (n-2)(n-1)/2

boditree
12-09-2009, 01:58 AM
Thanks, I'm glad there are smart people out there.