Between $1$ and $31$,$m$ numbers have been inserted in such a way that the resulting sequence is an $A.P.$ and the ratio of the $7^{\text{th}}$ and $(m-1)^{\text{th}}$ inserted numbers is $5:9$. Find the value of $m$.

If $a_1, a_2, a_3, \dots, a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$,then the value of $\sum_{r=1}^{21} a_r$ is:

Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be an $A$.$P$. of four terms such that each term of the $A$.$P$. and its common difference $l$ are integers. If $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=48$ and $\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}+l^{4}=361$,then the largest term of the $A$.$P$. is equal to

If the roots of the equation $6x^3-11x^2+6x-1=0$ are in harmonic progression,then the roots of $x^3-6x^2+11x-6=0$ will be in

Consider an $A$.$P$.: $a_1, a_2, \dots, a_n$,with $a_1 > 0$. If $a_2 - a_1 = -\frac{3}{4}$,$a_n = \frac{1}{4} a_1$,and $\sum_{i=1}^n a_i = \frac{525}{2}$,then $\sum_{i=1}^{17} a_i$ is equal to:

Let $\sum_{k=1}^{n} a_{k} = \alpha n^{2} + \beta n$. If $a_{10} = 59$ and $a_{6} = 7a_{1}$,then $\alpha + \beta$ is equal to: