1999 AHSME Problems/Problem 20
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Contents
Problem
The sequence satisfies , and, for all , is the arithmetic mean of the first terms. Find .
Solution 1
Let be the arithmetic mean of and . We can then write and for some .
By definition, .
Next, is the mean of , and , which is again .
Realizing this, one can easily prove by induction that .
It follows that . From we get that . And thus .
Solution 2
Let and . Then, , and so on.
Since , for all
Hence, We also know that
Subtracting from we get
~Benedict T (countmath1)
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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