Problem: Let a geometric progression with n terms have first term one, common ratio r and sum s, where r and s are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is
Answer Choices:
A. s1​
B. rns1​
C. rn−1s​
D. srn​
E. srn−1​
Solution:
The sum of the terms in the new progression is
1+r1​+…+rn−11​=rn−1rn−1+rn−2+…+1​=rn−1s​.