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.