Answer:

Explanation:

we have to find formula for nth term of series for this problem

Thus, we see series is 12, 9 ,6
Thus, we can see that series is decreasing by unit of 3.
and we know that when there is in increase or decrease in series by a a constant number then series is arithmetic progression series.
Hence, this is a decreasing AP series
In AP
nth term is given
nth term = a+(n-1)d
where a is the first term
d is the common difference
common difference is given by = nth term - (n-)th term
lets take 2nd and 1st term to get common difference.
d = 9-12 = -3
(note: this was obvious by looking at series itself that d is -3 as series was decreasing by 3 unit)
a = 12
thus, nth term = a +(n-1)d
nth term = 12 +(n-1)(-3)
nth term = 12 -3(n-1)
writing this in form of

Answer