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 Sequence of wrong means (Posted on 2011-05-26)
A) Start a sequence with any two real numbers, for the third and successive terms take the mean of all of the previous terms.

B) Same as part A) but take the mean incorrectly as follows: add the numbers together but instead of dividing by the number of terms, divide by one less than the number of terms.

C) Same as part A) and B) but take the mean incorrectly as follows: add the numbers together but instead of dividing by the number of terms, divide by one more than the number of terms.

What happens to the sequence in each case? Generalize further.

 See The Solution Submitted by Jer No Rating

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 generalized solution Comment 3 of 3 |
let first 2 real numbers are x and y: Case A) 3rd number would be (x+y)/2 -- call A(3), 4th would be: A(4) = (x + y + A(3)) / 3 = (x+y)/2 nth would be : (x + y + (n-3)*(x+y)/2)/(n-1) = (x+y)/2 so all subsequent number in this sequence would be mean of first 2 numbers. similary, case B) will have sum of first 2 numbers in all the subsequent places and case c) will have (x+y)/3 in one word, in all three cases 3rd number of the sequence will be at rest of the place... F(n) = F(3) --- n not equal to 1,2 more generalized solution: F(n,m) = (first number + second number)/m where n is position of sequence not equal to 1,2 and m is mean type (for case A it is 2, case B it's 1 while case C its 3)
 Posted by Ritesh on 2011-06-02 19:38:21

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