If $k+1, 8, k+9$ are three consecutive terms of an $A.P.$,then $k = \ldots$

The digits of a three-digit number form an $A.P.$ and their sum is $15$. The number obtained by reversing the order of the digits exceeds the original number by $594$. Find the original number.

Find the sum of those integers from $1$ to $500$ which are multiples of $2$ or $5$.

Which term of the $A.P.$ $85, 78, 71, \ldots$ is its first negative term? If the order of this term is $n$,find $S_{n}$.

For the $A.P.$ $\frac{3}{2}, \frac{7}{2}, \frac{11}{2}, \frac{15}{2}, \ldots,$ the common difference is

For the finite $A.P.$ $12, 21, 30, \ldots, 363,$ find the $12^{th}$ term from the end.