00:01
In this question, the relation given is a n equals to minus a n minus 1 plus 12 times of a n minus 2 and a not equals to 3.
00:15
A 0 3 and a 1 is 10 given.
00:18
First two terms are given.
00:20
Now, to find next term put n equals to 2.
00:25
So, a 2 will be equals to minus of a 2 minus 1 plus 12 times of a 2 minus 2.
00:34
So, a 2 equals to a 2 equals to minus of a 1 n plus 12 times of a 0.
00:44
So, a 2 equals to minus of a 1 which is 10 plus 12 times of a 0 which is 3.
00:54
So, a 2 equals to minus 10 plus 36.
01:00
So, this is equals to 26.
01:03
So, this implies the second term of the series will be 26.
01:10
Now, to find the third term put n equals to 3.
01:16
So, on substituting n equals to 3, a 3 equals to minus of a 3 minus 1 plus 12 times of a 3 minus 2.
01:29
So, a 3 will be equals to minus of a 2 plus 12 times of a 1.
01:38
So, a 2 is a minus 26 plus 12 times of a 1.
01:45
So, here a 1 that is a 10, a 1 is a 10.
01:49
So, 12 times of 10.
01:52
So, this is equals to minus 26 plus 120...