Question

The number of different ways in which the first $12$ natural numbers can be divided into three different groups such that numbers in each group are in A.P.

Answer 1

No group of four numbers from the first $12$ natural numbers can have the common difference $4$.
If one group including $1$ is selected with the common difference $1$, then the other two group can have the common difference $1$ or $2$.
Ex: $1^{\text{st}}=\{1,2,3,4\},2^{\text{nd}}=\{5,6,7,8\},3^{\text{rd}}=\{9,10,11,12\}$ or
$1^{\text{st}}=\{1,2,3,4)\},2^{\text{nd}}=\{5,7,9,11\},3^{\text{rd}}=\{6,8,10,12\}$

If one group including $1$ is selected with the common difference $2$, then one of the other two groups can have the common difference $2$ and the remaining group will have common difference $1$.
If one group including $1$ is selected with the common difference $3$, then the other two groups can have the common difference $3$.
Therefore, the required number of ways is $2+1+1=4$.