Question

Let the fourth term of an arithmetic progression be $6$ and the $m^{\text {th }}$ term be $18 .$ If $A.P$. has integral terms only, then the number of such $A.P$.s is ___

Answer 1

Given that $T_{4}=a+3 d=6$
and $T_{m}=a+(m-1) d=18$
$\Rightarrow (m-4) d=12$
Since terms are integers, we have
$(m-4)=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$
But $m > 0$
$\therefore (m-4)=\pm 1,\pm 2,\pm 3,-4,6,12$
So, $m$ has nine possible values.