the sum to infinity of a G.P. is thrice the first term. If the third term of the progression is 8, show that the sixth term is 64 over 27

Answer 1

The ratio, is r and the sum is a/(1-r).

Since a/(1-r) = 3a, r = 2/3

The answer to this question depends on how your series is numbered.

If you start with a₁ and the sequence is a_n = a₁ * r^n-1, then a₃ = a₁ * (2/3)² = 8

so that a₁ = 18.

Then a₆ = 18*(2/3)⁵ = 18*32/(27*9) = 64/27.

If you usually number the sequences starting with 0, the result may b different.