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Problems
Contests
International Contests
Austrian-Polish
2004 Austrian-Polish Competition
9
9
Part of
2004 Austrian-Polish Competition
Problems
(1)
Some average sequences
Source: Austrian-Polish 2004, Problem 9
7/5/2015
Given are the sequences
(
.
.
.
,
a
−
2
,
a
−
1
,
a
0
,
a
1
,
a
2
,
.
.
.
)
;
(
.
.
.
,
b
−
2
,
b
−
1
,
b
0
,
b
1
,
b
2
,
.
.
.
)
;
(
.
.
.
,
c
−
2
,
c
−
1
,
c
0
,
c
1
,
c
2
,
.
.
.
)
(..., a_{-2}, a_{-1}, a_0, a_1, a_2, ...); (..., b_{-2}, b_{-1}, b_0, b_1, b_2, ...); (..., c_{-2}, c_{-1}, c_0, c_1, c_2, ...)
(
...
,
a
−
2
,
a
−
1
,
a
0
,
a
1
,
a
2
,
...
)
;
(
...
,
b
−
2
,
b
−
1
,
b
0
,
b
1
,
b
2
,
...
)
;
(
...
,
c
−
2
,
c
−
1
,
c
0
,
c
1
,
c
2
,
...
)
of positive real numbers. For each integer
n
n
n
the following inequalities hold:
a
n
≥
1
2
(
b
n
+
1
+
c
n
−
1
)
a_n \geq \frac{1}{2} (b_{n+1} + c_{n-1})
a
n
≥
2
1
(
b
n
+
1
+
c
n
−
1
)
b
n
≥
1
2
(
c
n
+
1
+
a
n
−
1
)
b_n \geq \frac{1}{2} (c_{n+1} + a_{n-1})
b
n
≥
2
1
(
c
n
+
1
+
a
n
−
1
)
c
n
≥
1
2
(
a
n
+
1
+
b
n
−
1
)
c_n \geq \frac{1}{2} (a_{n+1} + b_{n-1})
c
n
≥
2
1
(
a
n
+
1
+
b
n
−
1
)
Determine
a
2005
a_{2005}
a
2005
,
b
2005
b_{2005}
b
2005
,
c
2005
c_{2005}
c
2005
, if
a
0
=
26
,
b
0
=
6
,
c
0
=
2004
a_0 = 26, b_0 = 6, c_0 = 2004
a
0
=
26
,
b
0
=
6
,
c
0
=
2004
.
Sequences
algebra