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

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?

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

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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

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

 

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. 

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.
­\(sum+\frac{1}{64}-\frac{1}{64}=­\)

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

­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

@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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