In a geometric progression, T2 = 24, T3 = 12(x+1), and S3 = 76. Find x.

Fresh day Ewurade, let's solve. Given: • Second term of a geometric progression (T_2) = 24 • Third term of a geometric progression (T_3) = 12(x+1) • Sum of the first three terms (S_3) = 76 Let the first term be a and the common ratio be r. The general formula for the n-th term of a geometric progression is T_n = ar^n-1. Step 1: Set up equations for the given terms. For the second term: T_2 = ar = 24 (1) For the third term: T_3 = ar^2 = 12(x+1) (2) Step 2: Find the common ratio (r) in terms of x. Divide equation (2) by equation (1): (ar^2)/(ar) = (12(x+1))/(24) r = (x+1)/(2) Step 3: Find the first term (a) in terms of x. From equation (1), a = (24)/(r). Substitute the expression for r: a = (24)/(x+1)2 a = (48)/(x+1) Step 4: Write the first three terms and use their sum. The first three terms are T_1 = a, T_2 = ar, and T_3 = ar^2. We are given T_1 + T_2 + T_3 = 76. Substitute the expressions for a, ar, and ar^2: (48)/(x+1) + 24 + 12(x+1) = 76 Step 5: Solve the equation for x. Subtract 24 from both sides: (48)/(x+1) + 12(x+1) = 76 - 24 (48)/(x+1) + 12(x+1) = 52 To simplify, let y = x+1. The equation becomes: (48)/(y) + 12y = 52 Multiply the entire equation by y (assuming y ≠ 0, which means x ≠ -1): 48 + 12y^2 = 52y Rearrange into a standard quadratic form: 12y^2 - 52y + 48 = 0 Divide by 4 to simplify the coefficients: 3y^2 - 13y + 12 = 0 Step 6: Solve the quadratic equation for y. Using the quadratic formula y = -b ± sqrt(b^2 - 4ac)2a: y = -(-13) ± sqrt((-13)^2 - 4(3)(12))2(3) y = 13 ± sqrt(169 - 144)6 y = 13 ± sqrt(25)6 y = (13 ± 5)/(6) This gives two possible values for y: y_1 = (13 + 5)/(6) = (18)/(6) = 3 y_2 = (13 - 5)/(6) = (8)/(6) = (4)/(3) Step 7: Find the corresponding values of x. Since y = x+1, then x = y-1. For y_1 = 3: x_1 = 3 - 1 = 2 For y_2 = (4)/(3): x_2 = (4)/(3) - 1 = (4)/(3) - (3)/(3) = (1)/(3) Both values are valid solutions. The value of x is 2 or (1)/(3). What's next?