The number of terms which are free from radical signs in the expansion of $(y^{1/5} + x^{1/10})^{55}$ is

Let the coefficients of powers of $x$ in the $2nd$, $3rd$, and $4th$ terms in the expansion of $(1+x)^{n}$, where $n$ is a positive integer, be in arithmetic progression. Then, the sum of the coefficients of odd powers of $x$ in the expansion is

If $a, b, c, d$ are any four consecutive coefficients of any expanded binomial,then $\frac{a + b}{a}, \frac{b + c}{b}, \frac{c + d}{c}$ are in

Let the coefficients of three consecutive terms $T_r$,$T_{r+1}$,and $T_{r+2}$ in the binomial expansion of $(a+b)^{12}$ be in a $G.P.$ and let $p$ be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $p+q$ is equal to:

If ${a_1}, {a_2}, {a_3}, {a_4}$ are the coefficients of any four consecutive terms in the expansion of ${(1 + x)^n}$,then $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

The number of terms in the expansion of $(y^{1/5} + x^{1/10})^{55}$,in which powers of $x$ and $y$ are free from radical signs are