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}$.