Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Help solving a geometric series problem?
Can someone clear something up for me please in this maths problem. (second term) T2= 0.5 and Sum to infinity(S) is 4. Find T1 and common ratio r. If I let T2 =ar^n-1 i end up with ar=0.5. Also S to infinity =a/1-r. Then I subbed in r=0.5/a into the equation for S and I got r = 4/3. Which is wrong. If I sub in a = 4(1-r) from the equation for S into T2 I get 0.25(2+2^.5) for r which is correct. Why can't I sub in the r=0.5/a
I'm sorry Kunal but I don't quite understand. If I sub in r= 0.5/a into that equation I send up with
a/(1-0.5a) =4 which means a= 4-2a which gives me 3a =4 which means a=4/3. How do I get the other value for a?
1 Answer
- 8 years agoFavorite Answer
when you substitute r=0.5/a in the other equation you will get two values of a and from one of those values you get r=4/3 which is obviously wrong as for an infinite GP r can not be greater than 1 and when you put the other value of a in the same equation where you got r=4/3 you will get the correct answer.