Difference between revisions of "1999 AHSME Problems/Problem 20"
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Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>. | Realizing this, one can easily prove by induction that <math>\forall n\geq 3;~ a_n=m</math>. | ||
| − | It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \boxed{179}</math>. | + | It follows that <math>m=a_9=99</math>. From <math>19=a_1=m-x</math> we get that <math>x=80</math>. And thus <math>a_2 = m+x = \(E)boxed{179}</math>. |
== See also == | == See also == | ||
Revision as of 21:44, 14 December 2018
Problem
The sequence 
statisfies 
, and, for all 
, 
is the arithmetic mean of the first 
terms. Find 
.
Solution
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 $a_2 = m+x = \(E)boxed{179}$ (Error compiling LaTeX. Unknown error_msg).
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. 
