$∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ =$

a^{2} + b^{2} + c^{2} - 3 a b c

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3 a b c

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3 a + 5 b

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0

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Solution

The correct option is D $0$
$∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ \begin{matrix} a + b & b & b a + 2 b & b & b a + 4 b & b & b \end{matrix} ∣ ∣ ∣ ∣$ $[\begin{matrix} C_{2}^{1} \to C_{2} \to C_{1} C_{3}^{1} \to C_{3} \to C_{2} \end{matrix}]$
$= 0$ Ans $[∵ C_{2} = C_{3}]$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a + b & a + 2 b & a + 3 b a + 2 b & a + 3 b & a + 4 b a + 4 b & a + 5 b & a + 6 b \end{matrix} ∣ ∣ ∣ ∣ =

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

(a^{2} + b^{2}) x^{2} - 2 b (a + c) x + (b^{2} + c^{2}) = 0

a, b, c

\frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a^{2} + b^{2} + c^{2} - a b - b c - c a}

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