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.
$\therefore b^{2}=a c, c^{2}=b d$
and $\frac{b}{a}=\frac{d}{c} \Rightarrow b c=a d \, ...(i)$
Now, $(a+b+c+d)^{2}=(a+b)^{2}+(c+d)^{2}$
$+2(a+b)(c+d)$
$=(a+b)^{2}+(c+d)^{2}+2(a c+a d+b c+b d)$
$=(a+b)^{2}+(c+d)^{2}+2\left(b^{2}+b c+b c+c^{2}\right)$
[from Eq. (i)]
$=(a+b)^{2}+(c+d)^{2}+2(b+c)^{2}$