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Given the formula S= a/ (1-r) find the value of r when a=3, and S=1 1/3 .

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Substituting a = 3 and S = 4/3 into the formula S = a / (1 - r) yields a common ratio (r) of -5/4 for the infinite geometric series.

The formula provided is for the sum (S) of an infinite geometric series with the first term (a) and the common ratio (r). To find the value of r when a = 3 and S = 4/3, you can substitute these values into the formula and solve for r:

S = a / (1 - r)

Given a = 3 and S = 4/3, the equation becomes:

(4/3) = (3 / (1 - r))

To solve for r, you can cross-multiply:

4(1 - r) = 3 * 3

Simplify the equation:

4 - 4r = 9

Now, isolate r by subtracting 4 from both sides:

-4r = 5

Finally, divide by -4 to solve for r:

r = -5/4

So, the value of r when a = 3 and S = 4/3 is -5/4.

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