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For every positive integer n, the nth term of sequence is given by an=

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For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Apr 2016, 11:26, edited 1 time in total.
Edited the question.

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inakihernandez wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/n+1. What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


Hi,
the moment you see the formula of nth number, it should tell you we are looking at a sequence ..
\(a_1 = \frac{1}{1}-\frac{1}{2}\)..
\(a_2= \frac{1}{2}-\frac{1}{3}\)..
\(a_3 = \frac{1}{3}-\frac{1}{4}\)..
.....
\(a_{99} = \frac{1}{99}-\frac{1}{100}\)..
\(a_{100} = \frac{1}{100}- \frac{1}{101}\)..
ADD all
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101} = 1-\frac{1}{101} = \frac{100}{101}\)
so when you add all terms, what is left is \(1-\frac{1}{101} = \frac{100}{101}\)..
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink]

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Hi chetan2u! thank you very much for your answer.

There is any way to calculate it using the formula (first term+last term)/2*number of terms?

because what I did as a first approach was a1= 1/2 ; a100= almost 0 => (1/2+0)/2*100= 25

I understood your solution but I was wondering if there is any way using the formula below, or was I using a wrong approach?

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inakihernandez wrote:
Hi chetan2u! thank you very much for your answer.

There is any way to calculate it using the formula (first term+last term)/2*number of terms?

because what I did as a first approach was a1= 1/2 ; a100= almost 0 => (1/2+0)/2*100= 25


No, don't use this formula here..
use it ONLY in an AP, where difference is common in consecutive numbers..
If you want to do approx, then too the terms should be atleast close to be called in AP
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink]

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inakihernandez wrote:
For every positive integer n, the nth term of sequence is given by an= 1/n - 1/(n+1). What is the sum of the first 100 terms?

(a) 1
(b) 0
(c) 25
(d) 99/100
(e) 100/101


For similar questions, check:
http://www.veritasprep.com/blog/2012/03 ... sequences/
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Re: gmat prep Q DS (series) [#permalink]

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New post 08 May 2016, 22:27
This clearly look like a GP series

a1 = 1 - 1/2 = 1/2

a2 = 1/2 - 1/3 = 1/6

r = 1/3

S100 = 1/2 (( 1 - (1/3)^100)/(1 - 1/3))

= 3/4 * (1 - 1/3)^100)

------> is this correct? how to solve beyond this to reach the ans?
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aniketm.87@gmail.com wrote:
For every positive integer n, the nth term of a sequence is given by an = 1/n - 1(n + 1). what is the sum of the 1st 100 terms in this sequence?

A) 0

B) 1/101

C) 99/100

D) 100/101

E) 1


hi
its not Gp..
At times we complicate the Q while simplifying it...

first write the numbers..
1st = \(1-\frac{1}{2}\)..
2nd = \(\frac{1}{2}-\frac{1}{3}\)
3rd = \(\frac{1}{3}-\frac{1}{4}\)..
99th = \(\frac{1}{99}-\frac{1}{100}\)..
1ooth = \(\frac{1}{100}-\frac{1}{101}\)...
add all ..
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101} = 1-\frac{1}{101} = \frac{100}{101}\)
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink]

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Bunuel, chetan2u,

For me, the tricky part here is that the GMAT assumes that we know that (1- 1/2) + (1/2 - 1/4) + (1/4 - 1/8) ... (1/99-1/100)+(1/100-1/101) = 1
How can we know that? Is there any specific rule?

Many thanks in advance,

Jaime

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Jaikuz wrote:
Bunuel, chetan2u,

For me, the tricky part here is that the GMAT assumes that we know that (1- 1/2) + (1/2 - 1/4) + (1/4 - 1/8) ... (1/99-1/100)+(1/100-1/101) = 1
How can we know that? Is there any specific rule?

Many thanks in advance,

Jaime


Hi,

Except the first and the last term, all the other terms get cancelled out. So the only calculation required here is 1 - 1/101.

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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink]

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New post 15 May 2016, 11:31
Hi Vyshak,

Can you please explain in more detail? I still don't see it.

Many thanks in advance.

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Jaikuz wrote:
Hi Vyshak,

Can you please explain in more detail? I still don't see it.

Many thanks in advance.


an= 1/n - 1/(n+1).

a1 = 1 - 1/2
a2 = 1/2 - 1/3
a3 = 1/3 - 1/4
.
.
.
a99 = 1/99 - 1/100
a100 = 1/100 - 1/101

Question: a1 + a2 + a3 + ...... + a99 + a100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......................... + 1/99 - 1/100 + 1/100 - 1/101 = 1 - 1/101 = 100/101

As you can see, except the first and the last terms the remaining terms get cancelled out.

Hope it helps.

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New post 30 Oct 2016, 10:25
Vyshak chetan2u : Why can't we use the Sum formula in this case? S100 = 100/2 * (2a + (100-1)d) ?

'Cuz if we do use, the answer comes to a value too far out from the answer choices, as per calculations.

Am i missing something?

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Re: For every positive integer n, the nth term of sequence is given by an= [#permalink]

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New post 30 Oct 2016, 10:41
Vishvesh88 wrote:
Vyshak chetan2u : Why can't we use the Sum formula in this case? S100 = 100/2 * (2a + (100-1)d) ?

'Cuz if we do use, the answer comes to a value too far out from the answer choices, as per calculations.

Am i missing something?


The sum formula that you have mentioned can be applied only when the sequence is in arithmetic progression. The given sequence is not in AP and hence cannot be applied.

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