If the coefficients of $x^{9}, x^{1} 0, x^{1} 1$ in the expansion of $(1 + x)^{n}$ are in arthimetic progression then $n^{2} - 4 l n$ =

398

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298

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-398

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198

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Solution

The correct option is C -398
$^{n} C_{9},^{n} C_{1} 0,^{n} C_{1} 1 ⟹ 2 = \frac{^{n} C_{9}}{^{n} C_{1} 0,} + \frac{^{n} C_{1} 1}{^{n} C_{1} 0,}$
$2 = \frac{10}{n - 9} + n - 10 11$
$⟹ 2 = \frac{110 + n^{2} - 19 n + 90}{11 n - 99}$
$⟹ n^{2} - 41 n = - 398$

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