If so then d = 12-4 = 8, a1 = 4

an = a1+(n-1)d

We see if 972 can be = an for some natural number n

972 = 4 + (n-1)(8)
972 = 4 + 8n - 8
972 = 8n - 4
976 = 8n
122 = n

Yes, since 122 is a natural number, and 972 is term number 122.

So

an = a1+(n-1)d  becomes

an = 4+(n-1)8 = 4+8n-8 = 8n-4

-----------------------------------------

If so then r = 12/4 = 3, a1 = 4

an = a1rn-1

We see if 972 can be = an for some natural number n

972 = 4·3n-1
Divide both sides by 4

243 = 3n-1

We can break 243 as 35

35 = sn-1

So the exponents of 3 must be equal:

5 = n-1`

6 = n

Yes, since 6 is a natural number.  In fact since 6 isn't so large,
we can write the sequence to 6 terms, though it isn't necessary,
it's just a check:

4, 12, 36, 108, 324, 972

So 972 really is term number 6.

an = a1rn-1  becomes

an = 4·3n-1

Edwin