Question

If $4$ numbers are to be selected from $1,2,3,......25$ such that they are be in AP, then number of ways of selecting numbers are

Answer 1

$a_n=a+(n-1)d$
$25=1+(4-1)d$
$d=8$
maximum difference between the numbers can be $8$.
If we take maximum difference $\left(8\right)$ then possible set of numbers are $(1,9,17,25)$
taking $d=7$ we get
$(1,8,15,22)$,$(2,9,16,23)$,$(3,10,17,24)$,$(4,11,18,25)$
similarly taking $d=6$ we get
$(1,7,13,19)$,$(2,8,14,20)$,$(3,9,15,21)$,.........$(7,13,19,25)$
every time reducing value of d with $1$ we get $3$ more set of numbers, so with $d=1$ we get $22$ set of numbers
So total number of ways of selecting numbers are
$22+19+16+......+1=92$