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,...

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

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,...

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...

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

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)

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

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.

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