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# how many minimum number of terms should be considered for

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Manager
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 157
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)
Followers: 1

Kudos [?]: 89 [0], given: 39

how many minimum number of terms should be considered for [#permalink]

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26 Jul 2014, 22:19
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25% (03:18) correct 75% (01:18) wrong based on 8 sessions

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how many minimum number of terms should be considered for below AP such that sum of the terms becomes less than zero

40,37,34,31,.......

a>26
b> 27
c>28
d>29
e>30
[Reveal] Spoiler: OA
Manager
Joined: 04 Sep 2012
Posts: 101
Location: Philippines
Concentration: Marketing, Entrepreneurship
Schools: Ross (Michigan) - Class of 2017
GMAT 1: 620 Q48 V27
GMAT 2: 660 Q47 V34
GMAT 3: 700 Q47 V38
GPA: 3.25
WE: Sales (Manufacturing)
Followers: 1

Kudos [?]: 40 [0], given: 504

Re: how many minimum number of terms should be considered for [#permalink]

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26 Jul 2014, 22:32
email2vm wrote:
how many minimum number of terms should be considered for below AP such that sum of the terms becomes less than zero

40,37,34,31,.......

a>26
b> 27
c>28
d>29
e>30

Number of Positive Numbers (+39 -- +3): $$40-1=\frac{39}{3}=13$$

Number of 0s: $$1$$

Number of Negative Numbers to counter Positive Numbers (-3 -- -39): $$13$$

Number of 3s needed to compensate the 1 point variance between (+39+1) & (-39-1) so that sum will be less than zero =$$1$$

$$13+1+13+1=28$$
Manager
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 157
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)
Followers: 1

Kudos [?]: 89 [0], given: 39

Re: how many minimum number of terms should be considered for [#permalink]

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27 Jul 2014, 01:10
email2vm wrote:
how many minimum number of terms should be considered for below AP such that sum of the terms becomes less than zero

40,37,34,31,.......

a>26
b> 27
c>28
d>29
e>30

sol:

S(n) = n/2 (2a +(n-1)d))<0

(we can remove 2 as this is positive and constant)

a=40 and d= -3

n(80+ (n-1) (-3))<0
n(83-3n)<0

if the co-efficient of the variable is negative then we can get negative sign (-1) out and flip the inequlaity sign

n(3n-83) >0

The value of N should be out side the boundaries

(+ve)-------0-----(-ve)--------83/3----(+ve)------

therefore n < 0 or n >83/3

number of values can not be less than 0 so we can discard that
so we remain with 83/3 = 27.66==>28
Re: how many minimum number of terms should be considered for   [#permalink] 27 Jul 2014, 01:10
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