What is the sum of all two-digit numbers which give a remainder of $4$ when divided by $6$?

If the $n^{th}$ term of an arithmetic progression is $\frac{(2n + 1)}{3}$,what is the sum of its first $19$ terms?

If ${a_1}, {a_2}, {a_3}, \dots, {a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$,then ${a_1} + {a_2} + {a_3} + \dots + {a_{23}} + {a_{24}} = $

If the sum of $n$ terms of an arithmetic progression is $3n^2 + 5n$ and $t_n = 164$,then $n = \dots$

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day,$4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is

If the sum of the $10$ terms of an $A.P.$ is $4$ times the sum of its $5$ terms,then the ratio of the first term to the common difference is: