00:01
So in this problem, we're first being asked to write an equation that will allow us to find the nth term for the arithmetic sequence where a sub 1 is 1 and d is negative 1 3rd.
00:11
Well, remember, our general formula is a sub n equals a sub 1 plus the quantity of n minus 1 times d.
00:20
So what we're going to do is substitute 1 in place of a sub 1 and negative 1 3 3 in place of d.
00:26
So when we do this, we get a sub n equals 1 plus the quantity of n minus 1 times negative 1 3rd.
00:35
And now we just have to simplify.
00:37
So to do this, we're going to distribute the negative 1 3rd to both terms in the preemphasy.
00:42
So we'll get a sub n equals 1 minus 1 3rd n plus 1 3rd.
00:49
Then we can combine our two like terms, while 1 plus 1 3rd is 1 in 1 3rd or simply 4 thirds...