If the coefficients of $2 n d, 3 r d$ and $4 t h$ terms in the expansion of ${(1 + x)}^{2 n}$ are in A.P., then

2 n^{2} - 5 n + 6 = 0

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n^{2} - 9 n + 3 = 0

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2 n^{2} - 3 n - 7 = 0

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2 n^{2} - 9 n + 7 = 0

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Solution

The correct option is C $2 n^{2} - 9 n + 7 = 0$
Here $r = 2$
For the terms to be in A.P
$(n - 2 r)^{2} = n + 2$
$(2 n - 4)^{2} = 2 n + 2$
$4 n^{2} + 16 - 16 n = 2 n + 2$
$4 n^{2} - 18 n + 14 = 0$
$2 n^{2} - 9 n + 7 = 0$

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