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A
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Difficulty:
35%
(medium)
Question Stats:
71% (01:57) correct
29%
(02:07)
wrong
based on 65
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\(\frac{1}{2}*\frac{1}{3}+\frac{1}{4}*\frac{1}{9}+\frac{1}{8}*\frac{1}{27}+...=\)
A) 0.2
B) 0.5
C) 0.75
D) 1.2
E) 2.5
A) 0.2
B) 0.5
C) 0.75
D) 1.2
E) 2.5
08 Feb 2022, 22:57
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gmatcafe
\(\frac{1}{2}*\frac{1}{3}+\frac{1}{4}*\frac{1}{9}+\frac{1}{8}*\frac{1}{27}+...=\)
\(\frac{1}{2}*\frac{1}{3}+\frac{1}{2^2}*\frac{1}{3^2}+\frac{1}{2^3}*\frac{1}{3^3}+...=\)
So, each successive term is multiplied by \(\frac{1}{2}*\frac{1}{3}\). Thus, it is a infinite GP.
DIRECT formula = \(\frac{a}{1-r}=\frac{\frac{1}{2}*\frac{1}{3}}{1-\frac{1}{2}*\frac{1}{3}}=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5}=0.2\)
Another method
Let \(x=\frac{1}{2}*\frac{1}{3}+\frac{1}{4}*\frac{1}{9}+\frac{1}{8}*\frac{1}{27}+...\)
\(\frac{1}{2}*\frac{1}{3}*x=\frac{1}{2}*\frac{1}{3}(\frac{1}{2}*\frac{1}{3}+\frac{1}{4}*\frac{1}{9}+\frac{1}{8}*\frac{1}{27}+...)\)
\(\frac{1}{6}*x=\frac{1}{4}*\frac{1}{9}+\frac{1}{8}*\frac{1}{27}+...=x-\frac{1}{2}*\frac{1}{3}=x-\frac{1}{6}\)
\(\frac{5x}{6}=\frac{1}{6}\)
\(x=\frac{1}{5}=0.2\)
A
23 Feb 2024, 12:27
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