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

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4 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
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4 years ago
```							Sorry. I got a problem in my internet connection. Remianing half of the solution is:-> X = 4 - 4r ( where -4
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4 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..
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4 years ago
```							HERE  X IS FIRST TERM                                                                                                                                               4 = X /  1 - r r = 1  – X/4 -1  0                                         THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD  THIS IS SHORT METHOD
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3 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 .....
```
3 years ago
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