Final answer:
To show that the ratio of the sum of the first n terms of a GP to the sum of terms from (n+1)th to 2nth term is 1/r^n, you use the GP sum formulas for the first n terms and first 2n terms, then subtract and simplify to find the desired ratio.
Step-by-step explanation:
The student is asking how to show that the ratio of the sum of the first n terms of a geometric progression (GP) to the sum of terms from the (n+1)th to the 2nth term is 1/rn, where r is the common ratio.
To prove this, we consider the formula for the sum of the first n terms of a GP (Sn): Sn = a(1 - rn)/(1 - r) where a is the first term of the GP.
For the sum of terms from (n+1)th to 2nth, this is equivalent to subtracting the sum of the first n terms from the sum of the first 2n terms.
The sum of the first 2n terms is: S2n = a(1 - r2n)/(1 - r).
The sum from (n+1)th to 2nth is: S2n - Sn.
We then divide the sum of the first n terms by this result to find the ratio:
(Sn) / (S2n - Sn) = a(1 - rn) / (1 - r) / (a(1 - r2n) / (1 - r) - a(1 - rn) / (1 - r)) = 1/rn.