An algebra problem by Keerthi Reddy
Given a , b and c follow an arithmetic progression , while a 2 1 , b 2 1 and c 2 1 also follow another arithmetic progression. Which of the following is true?
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2 solutions
this is the rigorous solution
It should be 1 / a 2 + 1 / c 2 = 2 / b 2
Simply, substitute that a=b=c=1 and solve..
it is non-rigorous
since a,b,c are in A.P. we have a-b=b-c. also 1 / a 2 , 1 / b 2 , 1 / c 2 are in A.P. so 1 / a 2 + 1 / c 2 = 1 / b 2
1 / a 2 + 1 / c 2 = 1 / b 2 , after simplifying becomes ( a b ) 2 + ( b c ) 2 = 2 ( a c ) 2 call this equation * now equation * can be written as ( a b ) 2 − ( a c ) 2 = ( a c ) 2 − ( b c ) 2 which can further be written as a 2 ( b + c ) ( b − c ) = c 2 ( a + b ) ( a − b ) . Since (a-b) = (b-c) we have a 2 ( b + c ) = c 2 ( a + b ) .
Q.E.D.