00:02
Okay, so we're asked to use our recursive formula, that's a .n is equal to a very common difference times a .n minus one.
00:13
To give two examples of geometric sequences whose third term is equal to 200.
00:20
Okay, so a3 is supposed to equal to 200.
00:29
Okay, so we have, let's see.
00:34
So i believe we can pick any value for a common ratio.
00:40
So let's let our first one equal to 2.
00:47
And an an minus 1 is equal to a .n.
00:53
Okay, so we know a3, which is 200, is equal to 2 times our a2.
01:06
Okay, so we can't really work with that.
01:07
We need to state our first value.
01:10
Okay, let's see.
01:14
So we know 2 times 100 is equal to 200.
01:20
So what if we let's have a 1.
01:22
Equal to 50, then a2 is equal to 2 times 50, which gives me 100, and then a 100 times 2 is equal to 200, that's a3.
01:34
So our first recursive formula is an, or a1 is equal to 50, and an is equal to 2 times a n minus 1 for n greater than or equal to 2...
