Best Answer

No it is not. It is most likely to be geometric

t(n) = 3^(n-1) for n = 1, 2, 3, 4.

But it is equally possible that the sequence is generated by the following order 4 polynomial:

t(n) = (15*n^4 - 118*n^3 + 381*n^2 - 494*n - 240)/24 for n = 1, 2, 3, 4.

Or infinitely many other polynomials.

Q: Is the pattern 1 3 9 27 arithmetic?

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U1 = 27 U{n+1} = U{n} - 3

what is the arthmetic sequence of 1 6 3 18 54 27 ? what is the missing terms ?

14

If you mean: 1 3 9 27 81 then it is 3*81 = 243

81 then 243

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30

U1 = 27 U{n+1} = U{n} - 3

xi+1 = 3 xi x1 = 1 x2 = 3 x 1 = 3 x3 = 3 x 3 = 9 x4 = 3 x 9 = 27 x5 = 3 x 27 = 81

what is the arthmetic sequence of 1 6 3 18 54 27 ? what is the missing terms ?

It is the start of an arithmetic sequence.

14

If you mean: 1 3 9 27 81 then it is 3*81 = 243

The pattern is the result of cubing 1 2 3 4 5 6 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 and so on

27 has more than two factors.Factors of 27 are 1, 3, 9 and 27.

I believe that the prime factors are 33. OR 3x3x3 which equals 27.

81 then 243

The divisors of 27 are: 1, 3, 9, 27.