1976 AMC 12 Problem 4

Question

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

\frac1s

\frac1{r^ns}

\frac{s}{r^{n-1}}

\frac{r^n}{s}

\frac{r^{n-1}}s

Write the reciprocal sum as

1+r^{-1}+\cdots+r^{-(n-1)}

Factor

r^{-(n-1)}

after reversing the order of the powers

Let

S'

be the reciprocal sum. Multiplying it by

r^{n-1}

reverses the original terms:

\begin{aligned} r^{n-1}S' &=r^{n-1}+r^{n-2}+\cdots+1\\ &=s. \end{aligned}

Hence

S'=\frac{s}{r^{n-1}}.

This also holds when

r=1.

Therefore, the correct answer is C.

Problem 4 in other years