Question

If $a, b, c, d$ are in $G.P$., then
$\frac{\left(a^{2} +b^{2} + c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)}{\left(ab+bc +cd\right)^{2}}= $

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

Answer: (A)

$b= ar, c= ar^{2}, d= ar^{3}$
$\therefore a^{2}+b^{2} +c^{2} = a^{2}\left(1+r^{2}+r^{4}\right)$
$ b^{2}+c^{2}+d^{2}= a^{2}r^{2} \left(1+r^{2}+r^{4}\right)$
$ab + bc +cd = a^{2}r \left(1+r^{2}+r^{4}\right)$
$ \therefore \frac{\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)}{\left(ab+bc+cd\right)^{2}} = 1$.