What is the missing number in the pattern 1, 3, 9,1,3,9, __ , 81,81?
2 Answers
Same thing as others wrote but written slightly differently.
The missing number is 27
Explanation:
It is a matter of spotting the relationship if you can. If not obvious you will need to experiment to find what type of sequence it is; geometric or arithmetic.
Observe that there is a connection of 3 in the numbers 1,3,9 in that
Lets experiment with that for a moment!
If we apply the same rule to 3 what will we get?
Ok. lets apply the same rule yet again!
Using 27, if we apply the rule once more and get 81 we have found the correct 'rule' for this sequence.
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This works so the missing number is 27
It would be
Explanation:
The normal assumption with this sequence is that it is arithmetic or geometric. The given numbers are consistent with a geometric sequence, resulting in the answer
However, note that we are not told that it is a geometric sequence, so there could be some other pattern.
For example, suppose it is a cubic sequence. Then taking the differences
Let us write out the given sequence, with
color(blue)(1), 3, 9, x, 811,3,9,x,81
Write out the sequence of differences between pairs of consecutive terms:
color(blue)(2), 6, x-9, 81-x2,6,x−9,81−x
Write out the sequence of differences of those differences:
color(blue)(4), x-15, 90-2x4,x−15,90−2x
Write out the sequence of differences of those differences:
color(blue)(x-19), 105-3xx−19,105−3x
If this is a constant sequence then we must have:
x-19 = 105-3xx−19=105−3x
Hence:
4x = 1244x=124
So:
x = 31x=31
Note that
Then the cubic formula for a term of the sequence can be written:
a_n = color(blue)(1)/(0!)+color(blue)(2)/(1!)(n-1)+color(blue)(4)/(2!)(n-1)(n-2)+color(blue)(12)/(3!)(n-1)(n-2)(n-3)an=10!+21!(n−1)+42!(n−1)(n−2)+123!(n−1)(n−2)(n−3)
color(white)(a_n) = 2n^3 - 10n^2 + 18n - 9an=2n3−10n2+18n−9
