The sum of an infinite geometric progression of positive terms is 4 . If the sum of the cubes of its terms is 192 7 , then the value of 64(a₂ + a₄) is equal to

Step-by-step solution

Let the first term of the infinite geometric progression be a and the common ratio be r . For the series to converge, we must have |r| 0 . The sum of the infinite G.P. is given by: a 1-r = 4 a = 4(1-r) The cubes of the terms form another infinite G.P. with first term a^3 and common ratio r^3 . Its sum is: a^3 1-r^3 = 192 7 Substitute a = 4(1-r) into this equation: 64(1-r)^3 (1-r)(1+r+r^2) = 192 7 Since r 1 , we can cancel (1-r) : 64(1-r)^2 1+r+r^2 = 192 7 (1-r)^2 1+r+r^2 = 3 7 Cross-multiplying gives: 7(1 - 2r + r^