How mamy geometric means are needed bdtween 2 and 1458 so that the sum of the resulting sepuence be 2186?

Answer 1

a = 2

2r^(n-1) = 1458
r^(n-1) = 729
729 = 9^3 = 3^6

recall that (r^n - 1) = (r-1)(1+r+r^2+...+r^(n-1))

so, our sum is a(r^n - 1)/(r-1)

2(9^4 - 1)/(9-1) = 2*6560/8 = 1640 nope
2(3^7 - 1)/(3-1) = 2*2186/2 = 2186
so we need 5 gm's between 2 and 1458

Check:

2,6,18,54,162,486,1458
sum = 2186