Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is square of an integer. The least possible value of the number of digits of $c$ is

The number of $5 -$tuples $(a, b, c, d, e)$ of positive integers such that

$I.$ $a, b, c, d, e$ are the measures of angles of a convex pentagon in degrees

$II$. $a \leq b \leq c \leq d \leq e$

$III.$ $a, b, c, d, e$ are in arithmetic progression is

Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$  If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals

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 $^n{C_4},{\,^n}{C_5},$ and ${\,^n}{C_6},$ are in $A.P.,$ then $n$ can be 

Show that the sum of $(m+n)^{ th }$ and $(m-n)^{ th }$ terms of an $A.P.$ is equal to twice the $m^{\text {th }}$ term.