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The sum of the first two terms of a GP is 5/3 and the sum of infinity of the series is 3. The common ratio is?
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The sum of the first two terms of a GP is 5/3 and the sum of infinity ...
Given:

- The sum of the first two terms of a GP is 5/3

- The sum of infinity of the series is 3

To find:

The common ratio (r)

Solution:

Step 1: Find the first two terms of the GP

Let the first term be a and the common ratio be r

Therefore, the second term will be ar

Given, the sum of the first two terms of the GP is 5/3

Therefore, a + ar = 5/3

=> a(1 + r) = 5/3

=> a = 5/3(1 + r)^-1


Step 2: Find the sum to infinity of the GP

The sum to infinity of a GP is given by S = a/(1 - r)

Given, the sum of infinity of the series is 3

Therefore, a/(1 - r) = 3

Substituting the value of a from step 1, we get:

5/3(1 + r)^-1 / (1 - r) = 3

=> 5/(3(1 + r)(1 - r)) = 3

=> 5 = 9(1 - r^2)

=> 5/9 = 1 - r^2

=> r^2 = 4/9

=> r = ±2/3


Step 3: Determine the value of r

The common ratio of a GP cannot be negative. Therefore, r = 2/3.

Answer:

The common ratio of the GP is 2/3.
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The sum of the first two terms of a GP is 5/3 and the sum of infinity of the series is 3. The common ratio is?
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