How with sequence questions

a)how many terms in the sequence: 9,12,15,..., (6p+15)?
b)insert 5 terms in the AP between -8 to 22

Thanks

a) I think it's infinity. Not much information is given.

b) Let the numbers inserted be a1, a2, a3, a4, a5.
So, the new AP is : -8, a1, a2, a3, a4, a5, 22.
This AP's 7th term is 22.
So, using general formula (a(n) = a + (n-1)d where a=first term, d=common difference, n=any term number of AP),
a(7) = 22 = (-8) + (7-1)d [Let d be common difference of AP]
--> 22 = -8 + 6d
--> d = 30/6 = 5.

So, common difference is 5.

Now, a(2) = a1 = -8 + 5 = -3.
a2 = -8 + 2(5) = +2
a3 = -8 + 3(5) = +7
a4 = -8 + 4(5) = +12
a5 = -8 + 5(5) = +17.