a,b,c,d are in continued proportion
⇒ab=bc=cd=k⇒a=bk,b=ck,c=dk,
To prove a2+b2+c2)(b2+c2+d2)=(ab+bc+cd+)2 substituting a,b,c
⇒((bk)2+(ck)2+(dk)2)(b2+c2+d2)=(b2k+c2k+d2k)2⇒k2(b2+c2+d2)(b2+c2+d2)=k2(b2+c2+d2)2⇒(b2+c2+d2)2=(b2+c2+d2)2
Hence proved.
If a, b, c, d are in GP, prove that
(a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2