Let x₁, x₂, x₃, be an arithmetic progression of real numbers with a positive common difference. If x₂ + x₈ = 7 and 3^ x₄ + 3^ x₆ = 252 , then the value of x₁₀ is equal to:

Step-by-step solution

Let the arithmetic progression have a common difference d > 0 . In an A.P., the sum of equidistant terms is constant. Therefore, x₄ + x₆ = x₂ + x₈ = 7 . We are given 3^ x₄ + 3^ x₆ = 252 . The product of these two terms is: 3^ x₄ 3^ x₆ = 3^ x₄ + x₆ = 3^7 = 2187 Thus, 3^ x₄ and 3^ x₆ are the roots of the quadratic equation: y^2 - 252y + 2187 = 0 (y - 9)(y - 243) = 0 Since the common difference d is positive, the A.P. is increasing, meaning x₄ This gives 3^ x₄ = 9 and 3^ x₆ = 243 . Therefore, x₄ = 2 and x₆ = 5 . The c