In a geometric progression, the sum to infinity is 12 and the sum of the first two terms is 20/3. Find the first term and the common ratio.

a) The first term and the common ratio of the progression. Step 1: Write down the formulas for the sum of the first n terms and the sum to infinity of a geometric progression. The sum of the first n terms is S_n = (a(1-r^n))/(1-r), where a is the first term and r is the common ratio. The sum to infinity is S_ = (a)/(1-r), valid for |r|<1. Step 2: Use the given information to form equations. Given that the sum to infinity is 12: S_ = (a)/(1-r) = 12 (1) Given that the sum of its first terms is (20)/(3). This refers to the sum of the first two terms, S_2. S_2 = a + ar = a(1+r) = (20)/(3) (2) Step 3: Solve the system of equations for a and r. From equation (1), express a in terms of r: a = 12(1-r) (3) Substitute equation (3) into equation (2): 12(1-r)(1+r) = (20)/(3) 12(1-r^2) = (20)/(3) Divide both sides by 12: 1-r^2 = (20)/(3 × 12) 1-r^2 = (20)/(36) Simplify the fraction: 1-r^2 = (5)/(9) Rearrange to solve for r^2: r^2 = 1 - (5)/(9) r^2 = (9-5)/(9) r^2 = (4)/(9) Take the square root of both sides: r = ±sqrt((4)/(9)) r = ±(2)/(3) The problem states that the geometric progression has positive terms. If r = -(2)/(3), the terms would alternate in sign (e.g., a, ar, ar^2, ). For all terms to be positive, r must be positive. Also, for S_ to exist, |r|<1, which is satisfied by both (2)/(3) and -(2)/(3). Therefore, we choose the positive value for r: r = (2)/(3) Now substitute r = (2)/(3) back into equation (3) to find a: a = 12(1-(2)/(3)) a = 12((3-2)/(3)) a = 12((1)/(3)) a = 4 Since a=4 and r=(2)/(3) are both positive, all terms in the progression will be positive. The first term is a=4 and the common ratio is r=(2)/(3). b) The sum of the first three terms Step 1: Use the values of a and r found in part (a). a=4 and r=(2)/(3). Step 2: Calculate the sum of the first three terms, S_3. Using the formula S_n = (a(1-r^n))/(1-r): S_3 = (4(1-(2)/(3))^3)1-(2)/(3) S_3 = (4(1-8)/(27))(1)/(3) S_3 = 4((27-8)/(27)) × 3 S_3 = 4((19)/(27)) × 3 S_3 = (4 × 19 × 3)/(27) S_3 = (4 × 19)/(9) S_3 = (76)/(9) The sum of the first three terms is (76)/(9).