If k is the sum of the reciprocals of the consecutive integers from 41 to 60 inclusive, which of the following are less than k?

Indicate all such statements.

❑ \frac{1}{4}

❑ \frac{1}{3}

❑ \frac{1}{2}


Since K is the sum of 1/41 to 1/60 and it is a evenly spaced set, we can easily calculate the sum of set k as follows:


(1/41 + 1/60)/(2) * 20 = 0,41

We multiply by 20 since we have 20 values in our set.

Hence A an B are the correct answers as they are < 0,41

I hope this helped !

S = 1/41 +1/42 + ... + 1/60
There are 40 elements >= 1/60 in that set => S > 20 * 1/60 = 1/3
Since 1/3 > 1/4 => S > 1/4 also

There are 40 elements < 1/40 in that set => S < 20 *1/40 = 1/2 => Eliminate C

The answer: A, B

Please provide OE.

Approximation works to solve this question.

We know that:

\(\frac{1}{40} > \frac{1}{41}, \frac{1}{42}, .... ,\frac{1}{49}\)

And:

\(\frac{1}{50} > \frac{1}{51}, \frac{1}{52}, .... ,\frac{1}{59}\)

So we can approximate \(k\):

\(k = \frac{1}{41} + \frac{1}{42} + ... + \frac{1}{51} + \frac{1}{52} + .... + \frac{1}{60}\)

\( k < \frac{1}{40} + \frac{1}{40} + .... + \frac{1}{50} + \frac{1}{50} + .... + \frac{1}{60}\)

There are 9 numbers between 41 and 49, and ten between 50 and 59, with the lone 60 at the end, so:

\( k < 9 * \frac{1}{40} + 10 * \frac{1}{50} + \frac{1}{60}\)

This simplifies to:

\( k < \frac{9}{40} + \frac{10}{50} + \frac{1}{60}\)

\( k < \frac{9}{40} + \frac{1}{5} + \frac{1}{60}\)

\(\frac{9}{40}\) is slightly less than \(\frac{10}{40}\) or \(\frac{1}{4}\), and \(\frac{1}{60}\) is less than \(\frac{1}{50}\), so essentially what we have is:

\( k < \frac{1}{4} + \frac{1}{5} + \frac{1}{50}\)

\( k < 0.25 + 0.2 + 0.02 \)

\( k < 0.47 \)

Giving us the answers A and B

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