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4

Part of 1976 AMC 12/AHSME

Problems(1)

Replacing terms of geometric progression with reciprocals

Source: 1976 AHSME Problem 4

5/14/2014
Let a geometric progression with nn terms have first term one, common ratio rr and sum ss, where rr and ss are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is
<spanclass=′latex−bold′>(A)</span>1s<spanclass=′latex−bold′>(B)</span>1rns<spanclass=′latex−bold′>(C)</span>srn−1<spanclass=′latex−bold′>(D)</span>rns<spanclass=′latex−bold′>(E)</span>rn−1s<span class='latex-bold'>(A) </span>\frac{1}{s}\qquad<span class='latex-bold'>(B) </span>\frac{1}{r^ns}\qquad<span class='latex-bold'>(C) </span>\frac{s}{r^{n-1}}\qquad<span class='latex-bold'>(D) </span>\frac{r^n}{s}\qquad <span class='latex-bold'>(E) </span>\frac{r^{n-1}}{s}
ratiogeometric sequenceAMC