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You've got a Harmonic Progression when the reciprocals of the values of your series became an Arithmetic Progression. In your case the AP is
32, 33, 34,....64

The only way I know to solve these is to find a range within which the sum will be. Here's how:

Find the total number of terms in your series. Here's how I do it:
tn = t1 + (n-1)d
64 = 32 + (n-1)1
n = 33

Add the first term and the last term to itself n number of times (33 here)
Thats 33(1/32) & 33(1/64)

So the sum of the series will be between 33/64 & 33/32.

i really wouldn't know how to solve this either, i'm curious to see the answer choices, but i would take this as a simple estimation problem

S=1/32 + 1/33 + ... +1/64

so you're adding a bunch of tiny fractions that are roughly equal to:

8/30 + 10/40 + 10/50 + 5/60

so you can add them rather quickly..

8/30 + 1/4 + 1/5 + 5/60 ->

4/15 + 1/4 + 3/15 + 5/60 ->

16/60 + 1/4 + 12/60 + 5/60 ->

33/60 + 1/4 ->

roughly 1/2 + 1/4 = 3/4

and since everything was rounded down i'd look for an answer a little higher. i would hope this could get me in the ballpark of the answers but you'd really have to see the answer choices to know for sure

.. and thats my very simple minded attempt to solve that =)

Very impressive Bluebird. You ask a good question.
My quant teacher told me there were different rules for AP's, GP's and HP's. He told me while we can get an exact value for the sum of an AP or a GP, we can only get a range within which the value of the sum of an HP series will be.

Maybe someone else might be able to add more to this but this is all I can tell you on this subject.

can you explain the bold part ? (1/32 + 1/64) => 3/64 ? the sum of this series cannot be greater than 1...a helpful hint: make sure you logically check your answer too..

Bluebird wrote:

Why can't we use the sum of a sequence equation?

n(a1+an)/2

where n=the number of values a1=the first value an=the last value

You've got a Harmonic Progression when the reciprocals of the values of your series became an Arithmetic Progression. In your case the AP is 32, 33, 34,....64

The only way I know to solve these is to find a range within which the sum will be. Here's how:

Find the total number of terms in your series. Here's how I do it: tn = t1 + (n-1)d 64 = 32 + (n-1)1 n = 33

Add the first term and the last term to itself n number of times (33 here) Thats 33(1/32) & 33(1/64)

So the sum of the series will be between 33/64 & 33/32.

thanks vikramjit_01 for explanation it's simple and makes sence