If the coefficients of $(3 r)^{t h}$ and $(r + 2)^{t h}$ terms in the expansion of $(1 + x)^{2 n}$ are equal $(r > 1, n > 2)$ , then

n = 2 r

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n = 3 r

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n = 2 r + 1

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n = 4 r

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Solution

The correct option is B $n = 2 r$
Consider the following
$T_{r + 1} =^{n} C_{r} a^{n - r} b^{r}$
Therefore the coefficient of $3 r^{t h}$ term is
$^{2 n} C_{3 r - 1}$ ...(i)
Therefore the coefficient of $(r + 2)^{t h}$ term is
$^{2 n} C_{r + 1}$ ...(ii)
Equating i and ii
$^{2 n} C_{3 r - 1} =^{2 n} C_{r + 1}$
Now $^{n} C_{r} =^{n} C_{n - r}$
Therefore
$3 r - 1 = 2 n - r - 1$
$4 r = 2 n$
$n = 2 r$
Hence answer is A

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