00:01
Hi there, so for this problem, we are, hi there, so for, so then first, we are given that for the first geometric progression, geometric progression, we have that first term is a, the common ratio is r, and the sum to infinity will be s.
00:39
Okay capital s.
00:41
Now for the second geometric progression we have that the first term is a.
00:52
The common ratio is capital r and the sum to infinity it is two times s.
01:03
The relationship between terms.
01:05
The third term of the first gp is equal to the second term of the second gp.
01:10
So with that said, let's solve our a in power of this problem.
01:15
Now for part a we are asked to show that r equals to two times capital r minus 1.
01:21
For this we start with for an infinity geometric series we know that x equals to a divided by 1 minus r.
01:28
Then for the other we will have 2 times s equals to a divided by 1 minus capital r.
01:35
Once we have this we need to express the relevant terms.
01:39
The general term of a gp of a geometric progression is that d n is equal to a times r elevated to n minus one for the third term in here we will have that this equals to i times r to the square the second term of the second gp will be a times capital r given that a times r square equals to a times capital r so we solve for art so so then r squared equals to capital r...