Question

In how many different ways can the first $12$ natural numbers be divided into three different groups such that numbers in each groups are in A.P.?

• A. $1$
• B. $5$
• C. $6$
• D. $4$

Answer 1

The correct option is D $4$

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 $2$.
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 $2+1+1=4$.