Let a₁, a₂, a₃, be G.P. of increasing positive terms such that a₂ a₃ a₄=64 and a₁+a₃+a₅= 813 7 . Then a₃+a₅+a₇ is equal to :

Step-by-step solution

Let first term be a and common ratio r > 1 . From a₂ a₃ a₄ = 64 : (ar)(ar^2)(ar^3) = a^3r^6 = 64 ar^2 = 4 . From a₁ + a₃ + a₅ = 813 7 : a(1 + r^2 + r^4) = 813 7 . Substituting a = 4 r^2 : 4 ( 1 r^2 + 1 + r^2 ) = 813 7 . Let u = r^2 : 28u^2 - 785u + 28 = 0 gives u = 28 (since r > 1 ). a₃ + a₅ + a₇ = ar^2(1 + r^2 + r^4) = 4(1 + 28 + 784) = 3252 .