Grade 11Algebra

The first term of an infinite geometric progression is X and it's sum is 4 .then sum of all integral values of X

Jaipreet

9 Years agoGrade 11

5 Answers

Faiz

9 Years ago

Let the common ratio be r.First term = X.Sum = 4 = X / ( 1 - r )In an infinite geometric progression the common ratio lies in the interval (0,1). This implies r belongs to (0,1).-> X = 4 - 4r ( where -4

Faiz

9 Years ago

Sorry. I got a problem in my internet connection. Remianing half of the solution is:-> X = 4 - 4r ( where -4

Faiz

9 Years ago

The web page is not accepting inequalities` sign like greater than or less than.-> X = 4 - 4r ( where -4r belongs to (-4,0) ). For X to be an integer -4r should be an integer possible integral values of -4r are -3,-2,-1.Hence possible values of X are 1,2,3. Summing all ans: 6..

prabhat

9 Years ago

HERE X IS FIRST TERM

4 = X / 1 - r

r = 1 – X/4

-1

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Naman

8 Years ago

Dear Student,Let the common ratio be rFirst term = XSum = 4 = X / ( 1 - r )[In an infinite geometric progression the common ratio lies in the interval (0,1).] This implies r belongs to (0,1) :- X = 4 - 4r ( where -4r belongs to (-4,0) ). For X to be an integer -4r should be an integer possible integral values of -4r are -3,-2,-1.Hence possible values of X are 1,2,3. Summing all ans: 6..Hope you understand the conceptThanks student .....