A geometric series has a positive common ratior. The series has a sum to infinity of 9 and the sum of the first 2 terms is 5 . Find the First 3 terms of the series

Step 1 ：Let the first term be \(a\) and the common ratio be \(r\). Since the sum to infinity is 9, we have \(\frac{a}{1-r} = 9\)

Step 2 ：The sum of the first 2 terms is 5, so \(a + ar = 5\)

Step 3 ：We are given that the common ratio \(r\) is positive, and we know that \(0 < r < 1\) for the sum to infinity to exist

Step 4 ：From the first equation, we can express \(a\) in terms of \(r\): \(a = 9(1-r)\)

Step 5 ：Substitute this expression for \(a\) into the second equation: \(9(1-r) + 9(1-r)r = 5\)

Step 6 ：Simplify the equation: \(9 - 9r + 9r - 9r^2 = 5\)

Step 7 ：Rearrange the equation into a quadratic: \(9r^2 - 9r + 4 = 0\)

Step 8 ：Factor the quadratic: \((3r - 1)(3r - 4) = 0\)

Step 9 ：Solve for \(r\): \(r = \frac{1}{3}\) or \(r = \frac{4}{3}\)

Step 10 ：Since \(0 < r < 1\), we choose \(r = \frac{1}{3}\)

Step 11 ：Substitute \(r\) back into the expression for \(a\): \(a = 9(1 - \frac{1}{3})\)

Step 12 ：Calculate \(a\): \(a = 9(\frac{2}{3}) = 6\)

Step 13 ：Now we can find the first 3 terms of the series: \(a, ar, ar^2\)

Step 14 ：Calculate the first 3 terms: \(6, 6(\frac{1}{3}), 6(\frac{1}{3})^2\)

Step 15 ：Simplify the first 3 terms: \(6, 2, \frac{2}{3}\)

Step 16 ：\(\boxed{6, 2, \frac{2}{3}}\)