Answer:
[tex]a_{n} = 12 -3(n-1)[/tex]
Step-by-step explanation:
[tex]a_{1} = 12\\a_{n} = a_{n-1} - 3[/tex]
we have to find formula for nth term of series for this problem
[tex]a_{1} = 12\\a_{2} = a_{2-1} - 3 = a_{1} -3 = 12 -3 = 9\\a_{3} = a_{3-1} - 3 = a_{2} -3 = 9 -3 = 6[/tex]
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 [tex]a_{n}[/tex]
[tex]a_{n} = 12 -3(n-1)[/tex] Answer
