Given an $A.P.$ whose terms are all positive integers. The sum of its first nine terms is greater than $200$ and less than $220$. If the second term in it is $12$, then its $4^{th}$ term is

Find the sum of all two digit numbers which when divided by $4,$ yields $1$ as remainder.

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

Find the $25^{th}$ common term of the following $A.P.'s$

$S_1 = 1, 6, 11, .....$

$S_2 = 3, 7, 11, .....$

If the sum of three numbers in $A.P.,$ is $24$ and their product is $440,$ find the numbers.

Let ${a_1},{a_2},.......,{a_{30}}$ be an $A.P.$, $S = \sum\limits_{i = 1}^{30} {{a_i}} $ and $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $.If ${a_5} = 27$ and $S - 2T = 75$ , then $a_{10}$ is equal to