Difference between revisions of "1999 AHSME Problems/Problem 20"
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Subtracting <math>a_1</math> from <math>198,</math> we get <math>b=a_2=\boxed{(E) 179}.</math> | Subtracting <math>a_1</math> from <math>198,</math> we get <math>b=a_2=\boxed{(E) 179}.</math> | ||
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+ | ~Benedict T (countmath1) | ||
== See also == | == See also == |
Revision as of 10:23, 17 April 2022
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
~Benedict T (countmath1)
See also
1999 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.