Thank you for reporting, we will resolve it shortly
Q.
If for positive integers $r > 1, n > 2$, the coefficients of the $(3r)^{th}$ and $(r + 2)^{th}$ powers of x in the expansion of $(1 + x)^{2n}$ are equal, then n is equal to
3r th term in the expansion of $(1+x)^{2 n}$
$={ }^{2 n} C _{3 r -1} x ^{3 r -1}$
and $(r+2)$ th term in the expansion of $(1+x)$
$=^{2 n } C _{ r +1} x ^{ r +1}$
Given that the binomial coefficients of $(3 r -1)$ and $( r +2)$ th terms are equal.
Thus ${ }^{2 n} C_{3 r-1}={ }^{2 n} C_{r+1}$
$
\Rightarrow 3 r -1= r +1
$
or $2 n=(3 r-1)+(r+1)$
$
\Rightarrow 2 r=2 \text { or } 2 n=4 r
$
$\Rightarrow r =1$ or $n=2 r$
But $r>1$
Therefore, n=2r