Question

If a b c d in continued proportion prove that (a-b/c + a-c/b)^2 - ( d-b/c + d-c/b)^2 = (a-d)^2 (1/c2 - 1/b2)​

Answer 1

Answer:

a, b, c, d are in continued proportion Rightarrow a b = b c = c d =k Rightarrow a=dk^ 3 , b = d * k ^ 2, c = dk LHS = ((a - b)/c + (a - c)/b) ^ 2 - ((d - b)/c + (d - c)/b) ^ 2 =( dk^ 3 -dk^ 2 dk + dk^ 3 -dk dk^ 2 )^ 2 -( d-dk^ 2 dk + d-dk dk^ 2 )^ 2 =(k^ 2 - 1 k )^ 2 -( 1 k^ 2 -k)^ 2 = (k^ 3 -1)^ 2 k^ 2 - (k^ 3 -1)^ 2 k^ 4 = (k^ 3 -1)^ 2 (k^ 2 -1) k^ 4 RHS = (a - d) ^ 2 * (1/(c ^ 2) - 1/(b ^ 2)) =(dk^ 3 -d)^ 2 ( 1 d^ 2 k^ 2 - 1 d^ 2 k^ 4 ) = (k^ 3 -1)^ 2 (k^ 2 -1) k^ 4 Hence, ((a - b)/c + (a - c)/b) ^ 2 - ((d - b)/c + (d - c)/b) ^ 2 = (a - d) ^ 2 * (1/(c ^ 2) - 1/(b ^ 2))