If a,b,c and d are in G.P., then the value of (a−c)2+(b−c)2+(b−d)2−(a−d)2 is
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Solution
Given : a,b,c and d are in G.P. So, ab=bc=cd⋯(1) Now, (a−c)2+(b−c)2+(b−d)2−(a−d)2=−2ac−2bc−2bd+2ad+2b2+2c2 Using equation (1), =−2b2−2bc−2c2+2bc+2b2+2c2=0