\((\frac{1}{7})+(\frac{1}{8})+(\frac{1}{9})\) is in between?
A. \((\frac{1}{6}) and (\frac{1}{5})\)
B. \((\frac{1}{5}) and (\frac{1}{4})\)
C. \((\frac{1}{4}) and (\frac{1}{3})\)
D. \((\frac{1}{3}) and (\frac{1}{2})\)
E. \((\frac{1}{2}) and 1\)
==>The sum of consecutive reciprocal numbers is decided by the first and the last number. In other words, you get
(1/9)+(1/9)+(1/9)< (1/7)+(1/8)+(1/9)< (1/7)+(1/7)+(1/7), and if you reorganize this, from
=1/3=3/9=(1/9)+(1/9)+(1/9)<(1/7)+(1/8)+(1/9)<(1/7)+(1/7)+(1/7)=3/7<3/6=1/2, you get between (1/3) and (1/2).
The answer is D.
Answer: D