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```				    Consider an infinite geometric series with first term and common ratio r. If its sum is 4 and the second term is 3/4, then : (A) a=4/7, r=3/7                                     (B) a =2, r = 3/8 (C) a = 3/2, r = 1/2                                 (D) a = 3, r = 1/4
```

7 years ago

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### Answers : (1)

```										Given, ar = 3/4 and a/(1-r) = 4
So, a = 3/4r. Putting this in a/(1-r) = 4,
we get 3/r(1-r) = 16, or 16r^2 -16r +3 = 0,=> r = 1/4 or 3/4.
Option (D) is the answer with a = 3 and r = 1/4.
```
7 years ago

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