If $a, b, c$ and $d$ are in $G.P.$ show that:
$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$
If $a, b, c$ and $d$ are in $G.P.$ Therefore,
$b c=a d$ ..........$(1)$
$b^{2}=a c$ .........$(2)$
$c^{2}=b d$ .........$(3)$
It has to be proved that,
$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$
$R.H.S.$
$=(a b+b c+c d)^{2}$
$=(a b+a d+c d)^{2}$ [ Using $(1)$ ]
$=[a b+d(a+c)]^{2}$
$=a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2}$
$=a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}\left(a^{2}+2 a c+c^{2}\right)$ [ Using $(1)$ and $(2)$ ]
$=a^{2} b^{2}+2 a^{2} c^{2}+2 b^{2} c^{2}+d^{2} a^{2}+2 d^{2} b^{2}+d^{2} c^{2}$
$=a^{2} b^{2}+a^{2} c^{2}+a^{2} c^{2}+b^{2} c^{2}+b^{2} c^{2}+d^{2} a^{2}+d^{2} b^{2}+d^{2} b^{2}+d^{2} c^{2}$
$=a^{2} b^{2}+a^{2} c^{2}+a^{2} d^{2}+b^{2} \times b^{2}+b^{2} c^{2}+b^{2} d^{2}+c^{2} b^{2}+c^{2} \times c^{2}+c^{2} d^{2}$
[ Using $(2)$ and $(3)$ and rearranging terms ]
$=a^{2}\left(b^{2}+c^{2}+d^{2}\right)+b^{2}\left(b^{2}+c^{2}+d^{2}\right)+c^{2}\left(b^{2}+c^{2}+d^{2}\right)$
$=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=$ $L.H.S$
$\therefore L .H.S. = R . H.S.$
$\therefore\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$
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