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Adding constant to geometric series

Source: 1959 AMC 12 Problem 12

August 13, 2013
ratiogeometric sequenceAMCgeometric seriesAMC 12algebra

Problem Statement

By adding the same constant to 20,50,10020,50,100 a geometric progression results. The common ratio is: <spanclass=′latex−bold′>(A)</span> 53<spanclass=′latex−bold′>(B)</span> 43<spanclass=′latex−bold′>(C)</span> 32<spanclass=′latex−bold′>(D)</span> 12<spanclass=′latex−bold′>(E)</span> 13 <span class='latex-bold'>(A)</span>\ \frac53 \qquad<span class='latex-bold'>(B)</span>\ \frac43\qquad<span class='latex-bold'>(C)</span>\ \frac32\qquad<span class='latex-bold'>(D)</span>\ \frac12\qquad<span class='latex-bold'>(E)</span>\ \frac13
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