All the integers from $1$ to $100$ are arranged in a $10 \times 10$ table as shown below. Prove that if some ten numbers are removed from the table, the remaining $90$ numbers contain 10 numbers in Arithmetic Progression. $1 \,\,\,\,2\,\, \,\,3 \,\,\,\,... \,\,10$ $11 \,\,12 \,\,13 \,\,... \,\,20$ $\,\,.\,\,\,\,.\,\,\,.$ $\,\,.\,\,\,\,.\,\,\,\,.$ $91 \,\,92 \,\,93\,\, ... \,\,100$