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

							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 
						

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

							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..
						

							
HERE  X IS FIRST TERM                                                                                                                                              
 4 = X /  1 - r
 r = 1  – X/4 
-1 
 0                                  
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							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 .....