Three consecutive terms of a geometric series are given in terms of a constant \( x \). \( U_{3}=(x+5), U_{4}=(4 x-1) \) and \( U_{5}=(2 x+3) \). Find the sum to infinity of the series. (Correct to four significant figures).

To solve for the sum to infinity of the geometric series given the terms \( U_{3}=(x+5), U_{4}=(4x-1) \), and \( U_{5}=(2x+3) \), we first need to find the common ratio \( r \) of the series. The common ratio \( r \) can be found using the formula for consecutive terms in a geometric series, given by: \[ r = \frac{U_{n+1}}{U_n} \] Thus, we can set up the following equations from our terms: 1. From \( U_3 \) and \( U_4 \): \[ r = \frac{U_4}{U_3} = \frac{4x - 1}{x + 5} \] 2. From \( U_4 \) and \( U_5 \): \[ r = \frac{U_5}{U_4} = \frac{2x + 3}{4x - 1} \] Setting these two expressions for \( r \) equal gives us the equation: \[ \frac{4x - 1}{x + 5} = \frac{2x + 3}{4x - 1} \] Cross-multiplying yields: \[ (4x - 1)^2 = (2x + 3)(x + 5) \] Expanding both sides: \[ 16x^2 - 8x + 1 = 2x^2 + 10x + 3x + 15 = 2x^2 + 13x + 15 \] Bringing all terms to one side results in: \[ 16x^2 - 8x + 1 - 2x^2 - 13x - 15 = 0 \] \[ 14x^2 - 21x - 14 = 0 \] Dividing the entire equation by 7 simplifies it: \[ 2x^2 - 3x - 2 = 0 \] Now we can factor this quadratic: \[ (2x + 1)(x - 2) = 0 \] Thus, we have two possible solutions: \[ x = -\frac{1}{2} \quad \text{or} \quad x = 2 \] Next, we need to calculate \( U_3, U_4, U_5 \) to determine the value of the common ratio \( r \) for valid \( x \). 1. If \( x = 2 \): \[ U_3 = 2 + 5 = 7, \quad U_4 = 4(2) - 1 = 7, \quad U_5 = 2(2) + 3 = 7 \] The ratio is: \[ r = \frac{7}{7} = 1 \] The series would not converge, so we discard this solution. 2. If \( x = -\frac{1}{2} \): \[ U_3 = -\frac{1}{2} + 5 = \frac{9}{2}, \quad U_4 = 4(-\frac{1}{2}) - 1 = -3, \quad U_5 = 2(-\frac{1}{2}) + 3 = 2 \] The ratio becomes: \[ r = \frac{U_4}{U_3} = \frac{-3}{\frac{9}{2}} = -\frac{6}{9} = -\frac{2}{3} \] To find the sum to infinity of a geometric series, the formula is: \[ S_{\infty} = \frac{U_3}{1 - r} \] Where \( r \) must be less than 1 in absolute terms: \[ S_{\infty} = \frac{\frac{9}{2}}{1 - (-\frac{2}{3})} = \frac{\frac{9}{2}}{1 + \frac{2}{3}} = \frac{\frac{9}{2}}{\frac{5}{3}} = \frac{9}{2} \times \frac{3}{5} = \frac{27}{10} = 2.7 \] Final answer: \[ \boxed{2.700} \] (to four significant figures).