The first term of an A.P. of 30 non-negative terms is $\frac{10}{3}$. If the sum of this A.P. is the cube of its last term, then its common difference is:

The number of ways, of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:

Let the smallest value of $k \in \mathbb{N}$, for which the coefficient of $x^3$ in $(1+x)^3+(1+x)^4+(1+x)^5+\ldots+(1+x)^{99}+(1+k x)^{100}, x \neq 0$, is $\left(43 n+\frac{101}{4}\right)\left({ }^{100} \mathrm{C}_3\right)$ for some $n \in \mathrm{~N}$, be $p$. Then the value of $p+n$ is :

Suppose that the mean and median of the non-negative numbers $21,8,17, a, 51,103, b, 13,67,(a>b)$, are 40 and 21 , respectively. If the mean deviation about the median is 26 , then $2 a$ is equal to :