Question

If $a,b,c$ and $d$ are distinct real numbers such that $(a^2+b^2+c^2)x^2-2x(ab+bc+cd)+(b^2+c^2+d^2)=0$, then;

• A. $a,b,c,d$ are in A.P.
• B. $a,b,c,d$ are in G.P.
• C. $a,b,c,d$ are in H.P.
• D. $ab=cd$

Answer 1

The correct option is B $a,b,c,d$ are in G.P.

$(a^2+b^2+c^2)x^2-2x(ab+bc+cd)+(b^2+c^2+d^2)=0$
We can simplify the expression by rearranging the terms as
$(ax-b{)}^2+(bx-c{)}^2+(cx-d{)}^2=0$
This will only be possible when
$ax=b,bx=c,cx=d$
$\Rightarrow d=cx=b{x}^2=a{x}^3$
Hence, $a,b,c,d$ are in a geometric progression with common ratio as $x$.