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?

Suppose $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of the first nine terms of the progression is $5:17$ and $110 < a_{15} < 120$,then the sum of the first ten terms of the progression is equal to -

What is the sum of all numbers between $100$ and $300$ that are divisible by $7$?

If $a, b, c, d, e, f$ are in $A.P.$,then the value of $e - c$ is

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,then find the number of terms.

The sum of the first $p, q,$ and $r$ terms of an $A.P.$ are $a, b,$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.