2 March 1814–3 June 1896 (Age 82)
South Carolina, United States
When Randolph Fuqua was born on 2 March 1814, in South Carolina, United States, his father, Randolph R Fuquay Jr., was 44 and his mother, Lydia Bomar, was 39. He married Mary A Hartzog on 8 March 1849, in Clayton, Barbour, Alabama, United States. They were the parents of at least 8 sons and 6 daughters. He lived in Barbour, Alabama, United States for about 6 years and Election Precinct 10 Coxs Mill, Barbour, Alabama, United States in 1880. He died on 3 June 1896, in Clayton, Barbour, Alabama, United States, at the age of 82, and was buried in Barbour, Alabama, United States.
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1814–1896Male
1831–1919Female
(14)
1850–1850Female
1851–1940Male
1852–1854Male
1855–1942Female
1857–1929Male
+9 More Children
1770–1840Male
1775–1821Female
(6)
1800–1860Male
1802–Male
1814–1896Male
1815–Male
1818–1916Female
+1 More Child
Dictionary of American Family Names © Patrick Hanks 2003, 2006.
Obituary for Ranolph Fuquay (Randall Fuqua)Clayton Courier, June 6, 1896Mr Randolph Fuquay died at his home, near Cox's Mill, on last Wednesday morning in the 80th year of his age. He was a good man a …
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