n=1 a(1) = 0.001
n=2 a(2) = 0.012
n=3 a(3) = 0.144
n=4 a(4) = 1.728
n=5 a(5) = 20.736
....
n=k a(k) = (12^(k-1))/1000
let n = k+1
a(k+1) = 12^(k)/1000
12(ak) = a(k+1)
12(12^k-1)/1000 = 12^k/1000
the 12 gets absorbed here.
12^(k-1+1)/1000 = 12^k/1000
Valid for k and k+1
therefore our equation
A(n) = 12^(n-1)/1000, n(greater than or equal to) 1
In order to answer the question is is necessary to know what the explicit formula was. But, since you have not bothered to provide that information, the answer is .
No.
a sequence of shifted geometric numbers
Since there is only one number, there is no sensible answer.
A descending geometric sequence is a sequence in which the ratio between successive terms is a positive constant which is less than 1.
Type yourWhich choice is the explicit formula for the following geometric sequence? answer here...
In order to answer the question is is necessary to know what the explicit formula was. But, since you have not bothered to provide that information, the answer is .
The sequence is neither arithmetic nor geometric.
None of the following could.
No.
Yes, that's what a geometric sequence is about.
This is not a geometric series since -18/54 is not the same as -36/12
a sequence of shifted geometric numbers
You mean what IS a geometric sequence? It's when the ratio of the terms is constant, meaning: 1, 2, 4, 8, 16... The ratio of one term to the term directly following it is always 1:2, or .5. So like, instead of an arithmetic sequence, where you're adding a specific amount each time, in a geometric sequence, you're multiplying by that term.
Since there is only one number, there is no sensible answer.
A geometric sequence is : a•r^n while a quadratic sequence is a• n^2 + b•n + c So the answer is no, unless we are talking about an infinite sequence of zeros which strictly speaking is both a geometric and a quadratic sequence.
antonette taño invented geometric sequence since 1990's