00:01
In this problem, we know we have an arithmetic sequence, and the eighth term, a sub 8, is 47, and the 21st term, a sub 21 is equal to 112.
00:09
So the first thing we're being asked to do is to find a sub 1 and d.
00:13
To do this, we're going to use our a sub n formula.
00:16
Remember, that's a sub n is equal to a sub 1 plus d times n minus 1.
00:22
So we're going to set up an equation for both of these two terms.
00:25
So for the first one, if a sub 8 is equal to 47, we know that n is equal to.
00:30
To 8.
00:31
So our equation would be 47 is equal to a sub 1 plus d times 8 minus 1.
00:37
Well, 8 minus 1 is 7.
00:39
So we'd have 47 equal to a sub 1 plus 70.
00:43
Now, we're going to do the same thing for our second term.
00:46
If a sub 21 is 112, then we know n is 21.
00:50
So we'd have 112 is equal to a sub 1 plus d times 21 minus 1.
00:57
21 minus 1 is 20.
00:59
So we have 112 equal to a sub 1 plus 20d.
01:03
And now we have a system of equations.
01:06
So i'm going to write them both together.
01:08
So 47.
01:09
And actually, you know what? i'm going to flip them around at the same time.
01:13
So we're going to have a sub 1 plus 7d equal to 47.
01:17
And a sub 1 plus 20d is equal to 112.
01:22
And now i'm going to solve by elimination...