Let a₁, a₂, a₃, be a geometric progression of increasing positive terms. If a₁ a₇ = 144 and a₂ + a₆ = 30 , then the sum of the squares of the first 6 terms of this progression is equal to:

Step-by-step solution

Let the geometric progression have first term a₁ and common ratio r . Since the terms are positive and increasing, r > 1 . The product of equidistant terms from the ends of a finite segment of a G.P. is constant. Therefore, a₂ a₆ = a₁ a₇ = 144 . We are given a₂ + a₆ = 30 . Thus, a₂ and a₆ are the roots of the quadratic equation: t^2 - 30t + 144 = 0 (t - 6)(t - 24) = 0 Since the G.P. is increasing, a₂ We know that a₆ = a₂ r^4 , so: 24 = 6 r^4 r^4 = 4 r^2 = 2 Also, a₂ = a₁ r a₂^2 = a₁^2 r^2 . Substituting the known v