Problem:
12 162 , 12 x , 12 y , 12 z , 12 1250 log 1 2 ​ 1 6 2 , log 1 2 ​ x , log 1 2 ​ y , log 1 2 ​ z , log 1 2 ​ 1 2 5 0
is an arithmetic progression. What is x x
Answer Choices:
A. 125 3 1 2 5 3 ​ 270 2 7 0 162 5 1 6 2 5 ​ 434 4 3 4 225 6 2 2 5 6 ​ Solution:
Because the terms form an arithmetic sequence,
12 y = 1 2 ( 12 162 + 12 1250 ) = 1 2 12 ( 162 ⋅ 1250 ) = 1 2 12 ( 2 2 3 4 5 4 ) = 12 ( 2 ⋅ 3 2 5 2 ) log 1 2 ​ y = 2 1 ​ ( log 1 2 ​ 1 6 2 + log 1 2 ​ 1 2 5 0 ) = 2 1 ​ log 1 2 ​ ( 1 6 2 ⋅ 1 2 5 0 ) = 2 1 ​ log 1 2 ​ ( 2 2 3 4 5 4 ) = log 1 2 ​ ( 2 ⋅ 3 2 5 2 ) ​
Then
12 x = 1 2 ( 12 162 + 12 y ) = 1 2 ( 12 ( 2 ⋅ 3 4 ) + 12 ( 2 ⋅ 3 2 5 2 ) ) = 1 2 12 ( 2 2 3 6 5 2 ) = 12 ( 2 ⋅ 3 3 5 ) = 12 270. log 1 2 ​ x = 2 1 ​ ( log 1 2 ​ 1 6 2 + log 1 2 ​ y ) = 2 1 ​ ( log 1 2 ​ ( 2 ⋅ 3 4 ) + log 1 2 ​ ( 2 ⋅ 3 2 5 2 ) ) = 2 1 ​ log 1 2 ​ ( 2 2 3 6 5 2 ) = log 1 2 ​ ( 2 ⋅ 3 3 5 ) = log 1 2 ​ 2 7 0 . ​
Therefore x = 270 x = 2 7 0 ​
OR
If ( B k ) = ( 12 A k ) ( B k ​ ) = ( log 1 2 ​ A k ​ ) d d ( A k ) ( A k ​ ) r = 1 2 d r = 1 2 d 162 , x , y , z , 1251 1 6 2 , x , y , z , 1 2 5 1 r r 1250 = 162 r 4 1 2 5 0 = 1 6 2 r 4 r = 5 3 r = 3 5 ​ x = 162 r = 162 ⋅ 5 3 = 270 x = 1 6 2 r = 1 6 2 ⋅ 3 5 ​ = 2 7 0 ​
The problems on this page are the property of the MAA's American Mathematics Competitions