If a, b, c and d are in G.P. show that

Solution:

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.

$\therefore\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$