gmihir wrote:

If S is the sum of reciprocals of a list of consecutive integers from 45 to 54, inclusive, S is approximately equal to

A. 0.1

B. 0.2

C. 0.3

D. 0.4

E. 0.5

Sorry, don't have an official answer.

Solved this using Bunuel's technique that I saw on another similar problem.

S = \(\frac{1}{45}+\frac{1}{46}+...+\frac{1}{54}\)

Get number of terms: \(54 - 45 + 1 = 10\)

Get lower limit: \(10*\frac{1}{54}=\frac{5}{27}=0.18\)

Get upper limit: \(10*\frac{1}{45}=\frac{2}{9}=0.22\)

\(0.18<S<0.22\)

Answer: B

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