If a, b, c and d are in G.P. show that $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2} .$
a, b, c, d are in G.P.
Therefore,
bc = ad … (1)
$b^{2}=a c \ldots(2)$
$c^{2}=b d \ldots(3)$
It has to be proved that,
$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c-c d)^{2}$
R.H.S.
$=(a b+b c+c d)^{2}$
$=(a b+a d+c d)^{2}[\cup \sin g(1)]$
$=[a b+d(a+c)]^{2}$
$=a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2}$
$=a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}\left(a^{2}+2 a c+c^{2}\right)$
$=a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}+d^{2} a^{2}+2 d^{2} b^{2}+d^{2} c^{2}[U \operatorname{sing}(1)$ and $(2)]$
$=a^{2} b^{2}+a^{2} c^{2}+a^{2} c^{2}+b^{2} c^{2}+b^{2} c^{2}+d^{2} a^{2}+d^{2} b^{2}+d^{2} b^{2}+d^{2} c^{2}$
[Using (2) and (3) and rearranging terms]
$=a^{2}\left(b^{2}+c^{2}+d^{2}\right)+b^{2}\left(b^{2}+c^{2}+d^{2}\right)+c^{2}\left(b^{2}+c^{2}+d^{2}\right)$
$=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)$
$=$ L.H.S.
$\therefore$ L.H.S. = R.H.S.
$\therefore\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$