Question

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

Answer 1

Let the common difference be $d$,
No set can have $d\ge 4$
As,
Element in each set is $4$, if $d=4$,
$1+3\times 4=13$
But we have to take upto $12$ natural numbers only.

Now,
$\text{Case}1:$
If the set containing $1$ has $d=1$,
$(1,2,3,4)$
Remaining numbers can be arranged in two ways,
$(5,6,7,8),(9,10,11,12)\text{and}(5,7,9,11),(6,8,10,12)$

$\text{Case}2:$
If the set containing $1$ has $d=2$,
$(1,3,5,7)$
Remaining numbers can be arranged in two ways,
$(2,4,6,8),(9,10,11,12)$

$\text{Case}3:$
If the set containing $1$ has $d=3$,
$(1,4,7,10)$
Remaining numbers can be arranged in two ways,
$(2,5,8,11),(3,6,9,12)$

Hence, total number of ways is $4.$