Let a₁, a₂, a₃, , a_ 2n be a geometric progression of positive terms. The sum of all the terms of this progression is 5 times the sum of its odd-positioned terms. If _ k=1 ^ 2n-1 k ₂ ( a_ k+1 a_k ) = 90 , then the value of n is equal to

Step-by-step solution

Let the common ratio of the geometric progression be r . The sum of all terms is given as S_ total = 5 S_ odd . We know that S_ total = S_ odd + S_ even . Substituting this, we get S_ odd + S_ even = 5 S_ odd S_ even = 4 S_ odd . In a geometric progression, the sum of the even-positioned terms is r times the sum of the odd-positioned terms. Therefore, S_ even = r S_ odd . Comparing the two equations, we find r = 4 . The ratio of any two consecutive terms a_ k+1 a_k is simply the common ratio r . Thus, ₂ ( a_ k+1 a_