If $(1 - x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . . . . + a_{2 n} x^{2 n}$ and $a_{0}, a_{1}, a_{2}, . . . . . . . . . . ., a_{2 n}$ are in arithmetic progression, then $a_{n}$ is

\frac{1}{n + 1}

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\frac{1}{n - 1}

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\frac{1}{2 n + 1}

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\frac{1}{2 n - 1}

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Solution

The correct option is C $\frac{1}{2 n + 1}$
Substituting $x = 1$ in the above mentioned binomial expression we get
$1 = a_{0} + a_{1} + a_{2} + . . . a_{n} + . . a_{2 n}$
Since they all are in A.P
$1 = (a_{0} + a_{2 n}) + (a_{1} + a_{2 n - 1}) + . . . a_{n}$
$= 1 = 2 a_{n} + 2 a_{n} + 2 a_{n} . . . a_{n}$
$1 = 2 n a_{n} + a_{n}$
$a_{n} (2 n + 1) = 1$
$a_{n} = \frac{1}{2 n + 1}$

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