If S is the sum of all the numbers of the form 1/n,

Question

If S is the sum of all the numbers of the form 1/n, where n is an integer from 33 to 64 inclusive, then S lies in which of the following intervals?

(A) 0 < S < 1/64
(B) 1/64 < S < 1/32
(C) 1/32 < S < 1/2
(D) 1/2 < S < 1
(E) 1 < S < 2

JeffTargetTestPrep · Accepted Answer

S = 1/33 + 1/34 + … + 1/64

We see that S is the sum of 64 – 33 + 1 = 32 different fractions. Of these fractions, 1/64 is the smallest, and 1/33 is the greatest.

Therefore,

32(1/64) < S < 32(1/33)

32/64 < S < 32/33 < 33/33

1/2 < S <1

Answer: D

gmatophobia · Answer

The number of terms between 33 and 64 both inclusive = (64 - 33) + 1 = 32

The middle term will of this sum of numbers will be approx  \frac{1}{50}

Approximate value of the sum =  \frac{1}{50} * 32   = 0.64

Hence, the value is greater than  \frac{1}{2}  and less than 1.

Option D

yipmewmew · Answer

I am curious to know can this way of taking the average and then multiplying to no of terms be applied to other questions too  gmatophobia ?

avinashnc · Answer

I would suggest a better way of thinking.
Given 32 terms each term 1/n satisfies the following condition
1/64<1/n<1/32
Now when added for 32 terms to get A
32*1/64<S<32*1/32 yielding option D

Posted from my mobile device

MclLaurent · Answer

Another approach

S = sum = (n (a1 + an))/2

(1) n = 64-33+1 (inclusive) = 33
(2) sum = (33(1/33 + 1/64)) / 2
= (33/33 + 33/64) / 2
= (1 + ~0,5) / 2
= ~1,5 / 2

S = ~0,75 = 3/4

Answer D

mbaaspirantss07 · Answer

gmatophobia  why is the middle term = 1/50? how did you reach this conclusion?

gmatophobia · Answer

33, 34, 35 ... 64 is in Arithmetic progress. We can find the middle term of an AP.

As the numbers are reciprocal here, the same concept applies.

P.S. If you like jargon, the terms are in Harmonic Progression.

nisen20 · Answer

Jeff has killed the game!

here's my clumsy attempt. hope it helps.

two ideas of how to estimate:

\(\frac{1}{33}+\frac{1}{34}>\frac{1}{17}\)
\(\frac{1}{32}+\frac{1}{33}<\frac{1}{16}\)

\(sum=\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}>\frac{1}{17}+...+\frac{1}{32}>\frac{1}{9}+...+\frac{1}{16}>\frac{1}{5}+...+\frac{1}{8}>\frac{1}{3}+\frac{1}{4}>\frac{1}{2}\)

by adding 1/64 and then subtracting it, we create the needed 1/32.
\(sum+\frac{1}{64}-\frac{1}{64}=\)

\(=\frac{1}{32}+\frac{1}{33}+...+\frac{1}{62}+\frac{1}{63}-\frac{1}{64}<\frac{1}{16}+...+\frac{1}{31}-\frac{1}{64}<\frac{1}{8}+...+\frac{1}{15}-\frac{1}{64}<\frac{1}{4}+...+\frac{1}{7}-\frac{1}{64}<\frac{1}{2}+\frac{1}{3}-\frac{1}{64}<1-\frac{1}{64}\)

\(\frac{1}{2}<sum<1-\frac{1}{64}\)

jack5397 · Answer

Hi Jeff,
I cannot understand how you arrived at S and the inequality which was created.
1/64 < S < 1/33 and then multiplying it with number of terms. Can you pls elaborate the approach or any concept hidden in this question.

TIA

GMATinsight · Answer

@ jack5397 
 
S = 1/33 + 1/34 + 1/35 + 1/36 + 1/37 + ..........+1/64 (total 32 terms)

Let's assume all terms (32 terms) in this series are all equal to the bigger (1/32) than biggest term (1/33) and call that series A
i.e. A = 1/32 + 1/32 + 1/32 + 1/32 + 1/32 + ..........+1/32 (32 terms) = 32((1/32) = 1

But since every term in this series is bigger than every term in series S therefore

S <  A

Also, Let's assume all terms (32 terms) in this series are all equal to the smallest term (i.e. 1/64) and call that series B
i.e. B = 1/64 + 1/64 + 1/64 + 1/64 + 1/64 + ..........+1/64 (32 terms) = 32((1/64) = 1/2

But since every term in this series is Smaller than every term in series S therefore

B  < S

Combining both the outcomes we get

B  < S <  A  
  1/2  < S <  1

Answer: Option D

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