Show that x+1&x+2&x+a +2&x+3&x+b +3&x+4&x+c =0 where a, b, c are … - Vidyadip

Question

Show that $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$ where $a, b, c$ are in A.P.

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Answer

$2\text{b}=\text{a}+\text{c}$
$\text{L.H.S}=\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix} [$Applying $R_2 = 2R_2]$
$=\frac{1}{2}\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\2\text{x}+4&2\text{x}+6&2\text{x}+2\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\0&0&0\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}$ $[\because2\text{b}=\text{a}+\text{c}]$
$[$Applying $R_2 → R_2 - (R_1 + R_3)]$
$=0$
$=\text{R.H.S}$

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