Question

If $a$, $b$, $c$, $d$ are in $G.P.$, then $(a+b+c+d{)}^2$ is equal to

• A. $(a+b{)}^2+(c+d{)}^2+2(b+c{)}^2$
• B. $(a+b{)}^2+(c+d{)}^2+2(a+c{)}^2$
• C. $(a+b{)}^2+(c+d{)}^2+2(b+d{)}^2$
• D. $(a+b{)}^2+(c+d{)}^2+(b+c{)}^2$
• E. $(a+b{)}^2+(c+d{)}^2+(b-c{)}^2$

Answer 1

Answer: (A)

Since, $a,b,c$ and $d$ are in GP.
$\therefore b^2=ac,c^2=bd$
and $\frac{b}{a}=\frac{d}{c}\Rightarrow bc=ad...(i)$
Now, $(a+b+c+d{)}^2=(a+b{)}^2+(c+d{)}^2$
$+2(a+b)(c+d)$
$=(a+b{)}^2+(c+d{)}^2+2(ac+ad+bc+bd)$
$=(a+b{)}^2+(c+d{)}^2+2\left(b^2+bc+bc+c^2\right)$
[from Eq. (i)]
$=(a+b{)}^2+(c+d{)}^2+2(b+c{)}^2$