if the fourth proportional of \({7}{{}}\frac{1}{5}\)\({4}{{}}\frac{1}{5}\)\({3}{{}}\frac{6}{7}\) is k, then \(\frac{4k+1}{4k-1}\) =

1
\(\frac{4}{5}\)
2
\(\frac{6}{5}\)
3
\(\frac{7}{5}\)
4
\(\frac{5}{4}\)

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