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)

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