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 $A.P.$

If $a, b, c$ are in $A.P.$,then $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ will be in

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

Let $s_1, s_2, s_3, \ldots, s_{10}$ be the sum of the first $12$ terms of $10$ arithmetic progressions whose first terms are $1, 2, 3, \ldots, 10$ and whose common differences are $1, 3, 5, \ldots, 19$ respectively. Then $\sum_{i=1}^{10} s_i$ is equal to

Four numbers are in arithmetic progression. The sum of the first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is