00:02
So for the first part, we are given two terms of an arithmetic sequence, the 121st, which is 47th, and the 146th term, which is negative 28.
00:27
So as we know for arithmetic sequence, the nth term is given by a .n equals a1 plus n minus 1 times d, where d is the common difference.
00:54
So we can write these two terms using this equation.
01:01
A121, which is equal to 47 is a.
01:07
Equal to a1 times 121 minus 1 times d.
01:17
So both a1 and d are unknown.
01:20
We can call this the first equation.
01:25
A146 equals minus 28 equals a1 again plus 146 minus 1 times d.
01:44
Just the second equation.
01:49
So if we subtract first equation by the second equation, we would get 47 minus 28, which is equal to 120d minus 145 d minus 145 d.
02:18
And this will be 75 equals minus 25d where now we can calculate d equals minus 3 and now we can substitute this to one of the equations and find a 1 so from equation 1 you can write 47 equals a 1 plus 122 times minus 3 so we get a1 equals 47 plus 360 which is equal to 407 and now we can calculate the calculate the 86 term a 86 which is equal to a1 plus 86 minus 1 times d and this is equal to 407 plus a times minus 3...