If $a, b, c$ and $d$ are in $G.P.$,show that:
$(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(ab+bc+cd)^{2}$

Since $a, b, c, d$ are in $G.P.$,we have $b=ar, c=ar^{2}, d=ar^{3}$.
$L.H.S. = (a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})$
$= (a^{2} + a^{2}r^{2} + a^{2}r^{4})(a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6})$
$= a^{2}(1 + r^{2} + r^{4}) \times a^{2}r^{2}(1 + r^{2} + r^{4})$
$= a^{4}r^{2}(1 + r^{2} + r^{4})^{2}$
$R.H.S. = (ab+bc+cd)^{2}$
$= (a(ar) + (ar)(ar^{2}) + (ar^{2})(ar^{3}))^{2}$
$= (a^{2}r + a^{2}r^{3} + a^{2}r^{5})^{2}$
$= (a^{2}r(1 + r^{2} + r^{4}))^{2}$
$= a^{4}r^{2}(1 + r^{2} + r^{4})^{2}$
Since $L.H.S. = R.H.S.$,the identity is proved.