1999 AHSME Problems/Problem 20
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.
It can be observed that for for all
Since We also know that
Subtracting from
we get
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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