a problem by aman
Algebra
Level
3
If a, b, c are 3 different numbers in A.P. then (a + 2b –c) (2b + c –a)(c + a –b) equals
abc
1/2abc
4abc
2abc
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1 solution
Since a , b and c are in A.P and let x be the common difference
Then, a = a , b = a + x and c = a + 2 x
Therefore, ( a + 2 b − c ) ( 2 b + c − a ) ( c + a − b ) = ( a + 2 ( a + x ) − ( a + 2 x ) ) ( 2 ( a + x ) + ( a + 2 x ) − a ) ( ( a + 2 x ) + a − ( a + x ) ) = ( a + 2 a + 2 x − a − 2 x ) ( 2 a + 2 x + a + 2 x − a ) ( a + 2 x + a − a − x ) = ( 2 a ) ( 2 a + 4 x ) ( a + x ) = ( 2 a ) ( 2 ( a + 2 x ) ) ( a + x ) = ( 2 a ) ( 2 c ) ( b ) = 4 a b c