Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.

If the first term of an $A.P. $ be $10$, last term is $50$ and the sum of all the terms is $300$, then the number of terms are

Given that $n$ A.M.'s are inserted between two sets of numbers $a,\;2b$and $2a,\;b$, where $a,\;b \in R$. Suppose further that ${m^{th}}$ mean between these sets of numbers is same, then the ratio $a:b$ equals

The sixth term of an $A.P.$ is equal to $2$, the value of the common difference of the $A.P.$ which makes the product ${a_1}{a_4}{a_5}$ least is given by

If the sum and product of the first three term in an $A.P$. are $33$ and $1155$, respectively, then a value of its $11^{th}$ tern is

Let $s _1, s _2, s _3, \ldots \ldots, s _{10}$ respectively be the sum to 12 terms of 10 A.P.s whose first terms are $1,2,3, \ldots, 10$ and the common differences are $1,3,5, \ldots, 19$ respectively. Then $\sum \limits_{i=1}^{10} s _{ i }$ is equal to