Let fourth term of an arithmetic progression be 6 and m^th term be 18. If A. P. has integral terms only then the numbers of such A.P.s is _________.

Given that T₄ = a + 3d = 6

and ${\mathrm{T}}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1)\mathrm{d}=18$

$\Rightarrow (\mathrm{m}-4)\mathrm{d}=12$

Since terms are integers, we have

$(\mathrm{m}-4)=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

But m > 0.

$\therefore (\mathrm{m}-4)=\pm 1,\pm 2,\pm 3,4,6,12$.

So, m has nine Possible values.