k method
if a:b=c:d prove that
(abcd)(a-²+b-²+c-²+d-²)=(a²+b²+c²+d²)
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Given a,b,c,d are in continued proportion
⟹ba=cb=dc=k(say)
⟹c=dk,b=ck=k2d,a=bk=k3d
LHS=(a2−b2)(c2−d2)=(k6d2−k4d2)(k2d2−d2)=k4d4(k2−1)2=(k4d2−k2d2)2=((k2d)2−(kd)2)2
=(b2−c2)2=RHS
Hence Proved
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