00:01
Suppose we have a sequence where the second term is 7, and the fourth term is 1 ,575.
00:06
Let's determine the nth term.
00:08
So we have a formula from the book stating that the nth term, a .n, is equal to a1, the first term, times r, the common ratio, raised the n minus 1 power.
00:17
So to solve this, we need to figure out a1 and r.
00:21
Let's try r first.
00:22
So let's talk about the definition of common ratio.
00:25
That is, if you multiply some term, let's say a2 by r, you will get the next term.
00:32
So in this case, a3.
00:34
Then if you multiply a3 by r, you will get the next term up, a4.
00:39
That's just how common ratio works.
00:41
Multiplying any term by r will give you the next term up in line.
00:47
So let's try multiplying both sides of this equation by r.
00:50
That will give us a2 times r times r equals a3 times r.
00:58
But a3 times r is a4, as previously discussed.
01:03
So a2 times r squared equals a4...