If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in

H.P.

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G.P.

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A.P.

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A.G.P

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Solution

The correct option is C A.P.
Since the given lines are concurrent
$∴ ∣ ∣ ∣ ∣ \begin{matrix} a & 12 & 1 b & 13 & 1 c & 14 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} a & 12 & 1 b - a & 1 & 0 c - b & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$[A p p l y i n g R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1}] \Rightarrow b - a - c + b = 0 o r 2 b = a + c \Rightarrow a, b, c a r e i n A . P .$

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Similar questions

a x + b y + c = 0,

b x + c y + a = 0

c x + a y + b = 0

If a, b, c are in A.P., prove that the straight lines $a x + 2 y + 1 = 0, b x + 3 y + 1 = 0$ and $c x + 4 y + 1 = 0$ are concurrent.

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