Let there be four distinct integers in an increasing arithmetic progression. One of these integers is equal to the sum of the squares of the other three. What is the product of all these numbers?

Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$

Different $A.P.$s are constructed with the first term $100$,the last term $199$,and integral common differences. The sum of the common differences of all such $A.P.$s having at least $3$ terms and at most $33$ terms is:

If the sum of $n$ terms of the sequence $2, 5, 8, 11, \dots$ is $60100$,then $n = \dots$

Let $A_1, A_2, \dots, A_{39}$ be $39$ arithmetic means between the numbers $59$ and $159$. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to:

If $x, y, z$ are in an arithmetic progression,and $a$ is the arithmetic mean of $x$ and $y$,and $b$ is the arithmetic mean of $y$ and $z$,then what is the arithmetic mean of $a$ and $b$?