M is the sum of the reciprocals

[rajeet123]

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

[rajeet123]

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Given that M=1201+1202+1203+...+1300M=1201+1202+1203+...+1300. Notice that 1/201 is the larges term and 1/300 is the smallest term.

If all 100 terms were equal to 1/300, then the sum would be 100/300=1/3, but since actual sum is more than that, then we have that M>1/3.

If all 100 terms were equal to 1/200, then the sum would be 100/200=1/2, but since actual sum is less than that, then we have that M<1/2.

Therefore, 1/3<M<1/2.

Answer: A.

[Scott@TargetTestPrep]

M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?

(A) 1/3 < M < 1/2
(B) 1/5 < M < 1/3
(C) 1/7 < M < 1/5
(D) 1/9 < M < 1/7
(E) 1/12 < M < 1/9

Let's first analyze the question. We are trying to find a potential range for M in which M is the sum of the reciprocals from 201 to 300, inclusive. Thus, M is:

1/201 + 1/202 + 1/203 + ... + 1/300

Since we probably would not be expected to do such time-consuming arithmetic (i.e., to add 100 fractions, each with a different denominator), that is exactly why the answer choices are in the form of an inequality. Thus, we do not need to know the EXACT value of M. The easiest way to determine the RANGE of values for M is to use easy numbers that can be quickly manipulated.

Notice that 1/200 is greater than each of the addends and that 1/300 is less than or equal to each of the addends. Therefore, instead of trying to add together 1/201 + 1/202 + 1/203 + ... + 1/300, we are going to add 1/200 one hundred times and 1/300 one hundred times. These two sums will give us a high estimate of M and a low estimate of M. Again, we are adding 1/200 one hundred times and 1/300 one hundred times because there are 100 numbers from 1/201 to 1/300.

Instead of actually adding each one of these values one hundred times, we will simply multiply each value by 100:

1/300 x 100 = 1/3 . This value is the low estimate of M.

1/200 x 100 = 1/2. This value is the high estimate of M.

We see that M is between 1/3 and 1/2.

Answer: A