Question

If $3x^2-2(a-d)x+(a^2+2(b^2+c^2)+d^2)=2(ab+bc+cd)$ , then ?

• A. a, b, c, d are in G .P.
• B. a, b, c, d are in H .P.
• C. a, b, c, d are in A .P.
• D. None of these

Answer 1

Answer: (C)

a, b, c, d are in A .P.

Given equation is $3x^2-2(a-d)x+(a^2+2(b^2+c^2)+d^2)=2(ab+bc+cd)$

We can split the inside bracket as:

$3x^2-2(a-d)x+\left[\right(a^2+b^2-2ab)+(b^2+c^2-abc)+(c^2+d^2-2cd\left)\right]=0$

$3x^2-2(a-d)x+\left[\right(a-b{)}^2+(b-c{)}^2+(c-d{)}^2]=0$ .....(i)

Now differentiating equation (i) w.r.t. $x$, we get

$6x-2(a-d)=0$

$\Rightarrow 6x=2(a-d)$

$\Rightarrow x=\frac{(a-d)}{3}$

Substituting this value of $x$ in equation (i), we get

$3{\left[\frac{(a-d)}{3}\right]}^2-2(a-d)\left[\frac{(a-d)}{3}\right]+\left[\right(a-b{)}^2+(b-c{)}^2+(c-d{)}^2]=0$

$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-d{)}^2=\frac{(a-d{)}^2}{3}$

$\Rightarrow 3\left[\right(a-b{)}^2+(b-c{)}^2+(c-d{)}^2]=(a-d{)}^2$ ....(ii)

This equation is satisfied only when $a,b,c,d$ are in A.P.