Let $a_1, a_2 , a_3,.....$ be an $A.P$, such that $\frac{{{a_1} + {a_2} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + ..... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}};p \ne q$. Then $\frac{{{a_6}}}{{{a_{21}}}}$ is equal to

The sums of $n$ terms of three $A.P.'s$ whose first term is $1$ and common differences are $1, 2, 3$ are ${S_1},\;{S_2},\;{S_3}$ respectively. The true relation is

The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6 .$ Find the ratio of their $18^{\text {th }}$ terms.

If the sum of $n$ terms of an $A.P$. is $2{n^2} + 5n$, then the ${n^{th}}$ term will be

The arithmetic mean of the nine numbers in the given set $\{9,99,999,...., 999999999\}$ is a $9$ digit number $N$, all whose digits are distinct. The number $N$ does not contain the digit

If ${S_1},\;{S_2},\;{S_3},...........{S_m}$ are the sums of $n$ terms of $m$ $A.P.'s$ whose first terms are $1,\;2,\;3,\;...............,m$ and common differences are $1,\;3,\;5,\;...........2m - 1$ respectively, then ${S_1} + {S_2} + {S_3} + .......{S_m} = $