If the coefficients of $r^{th}$ and ${(r + 1)}^{th}$ terms in the expansion of ${(1 + x)}^{24}$ are in the ratio 12 : 13, then r is the root of the quadratic equation

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$x^{2} - 5 x + 6 = 0$

$x^{2} - 11 x + 30 = 0$

$x^{2} - 14 x + 13 = 0$

$x^{2} - 14 x + 24 = 0$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \frac{T_{r}}{T_{r + 1}} & = & \frac{12}{13} \Rightarrow \frac{{}^{24}{C_{r - 1}}}{{}^{24}{C_{r}}} = \frac{12}{13} \\ \frac{13}{12} & = & \frac{24 - r + 1}{r} \\ 13 r & = & 300 - 12 r \\ 25 r & = & 300 \\ r & = & 12 \end{array}$

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