MCQ
If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha - b}\\b&c&{b\alpha - c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then
  • A
    $a,b,c$ are in $A. P.$
  • $a,b,c$ are in $G. P.$
  • C
    $a,b,c$ are in $H. P.$
  • D
    None of these

Answer

Correct option: B.
$a,b,c$ are in $G. P.$
b
(b) $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha - b}\\b&c&{b\alpha - c}\\2&1&0\end{array}\,} \right| = 0$

==> $a[ - (b\alpha - c)] - b[ - 2(b\alpha - c)] + [a\alpha - b)(b - 2c)] = 0$

==>$ - ab\alpha + ac + 2{b^2}\alpha - 2bc + ab\alpha - 2ac\alpha - {b^2} + 2bc = 0$

==> $ac + 2{b^2}\alpha - 2ac\alpha - {b^2} = 0$

==> $(ac - {b^2}) - 2\alpha (ac - {b^2}) = 0$

==> $ac - {b^2} = 0$or $1 - 2\alpha = 0$ $ \Rightarrow $ ${b^2} = ac$   or $\alpha = \frac{1}{2}$

(As given in question)

So, ${b^2} = ac$ i.e, $a,b,c$ are in $G.P.$

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