Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 x)^{ n }$ be in the ratio $2: 5: 8$. Then the coefficient of the term, which is in the middle of these three terms, is $...........$.

Let for the $9^{\text {th }}$ term in the binomial expansion of $(3+6 x)^{n}$, in the increasing powers of $6 x$, to be the greatest for $x=\frac{3}{2}$, the least value of $n$ is $n_{0}$. If $k$ is the ratio of the coefficient of $x ^{6}$ to the coefficient of $x ^{3}$, then $k + n _{0}$ is equal to.

If the coefficents of ${x^3}$ and ${x^4}$ in the expansion of  $\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$ in powers of $x$ are both zero, then $ (a,b) $ is equal to 

Coefficient of $t^{20}$ in the expansion of $(1 + t^2)^{10}(1 + t^{10})(1 + t^{20})$ is

The coefficient of $x^{37}$ in the expansion of $(1-x)^{30} \, (1 + x + x^2)^{29}$ is :

The coefficients of three successive terms in the expansion of ${(1 + x)^n}$ are $165, 330$ and $462$ respectively, then the value of n will be