Question

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

• A. ${\left(c+d\right)}^2$
• B. ${\left(c-d\right)}^2$
• C. ${\left(a+d\right)}^2$
• D. ${\left(a-d\right)}^2$

Answer 1

Answer: (D)

Since $b^2=ca$,
$c^2=bd$,
$bc=ad$
$\therefore {\left(b-c\right)}^2+{\left(c-a\right)}^2+{\left(b-d\right)}^2$
$=b^2-2bc+c^2-2ca+a^2+b^2-2bd+d^2$
$=a^2+d^2-2bc$
$=a^2+d^2-2ad$
$={\left(a-d\right)}^2$.