A sequence of squares S₁, S₂, S₃, is constructed such that their side lengths form a strictly increasing geometric progression. If the product of the perimeters of S₁, S₃ , and S₅ is 4096 , and the sum of the lengths of the diagonals of S₂ and S₄ is 10 2 , then the area of the square S₆ is equal to

Step-by-step solution

Let the side length of the n -th square S_n be a_n . The sequence a₁, a₂, a₃, forms a strictly increasing geometric progression with common ratio r > 1 . The perimeter of S_n is P_n = 4a_n . The product of the perimeters of S₁, S₃ , and S₅ is given as 4096 : (4a₁)(4a₃)(4a₅) = 4096 64(a₁ a₃ a₅) = 4096 a₁ a₃ a₅ = 64 Using the property of a geometric progression, a₁ a₅ = a₃^2 . Thus: a₃^3 = 64 a₃ = 4 The length of the diagonal of S_n is D_n = a_n 2 . The sum of the diagonals of S₂ and S₄ is 10 2 : a₂ 2 + a₄ 2 = 10 2 a