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by clock60 » Thu Dec 23, 2010 7:18 am
hi guys please share your ideas about this problem

The sum of first N-1 terms of AP is either zero or positive. but sum of first N terms is negative. What is the value of N?

A) Common difference is -4 and 7th term is last positive term
B) 25 is first term and there are 7 positive terms and all are integers
ao is B
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by anshumishra » Thu Dec 23, 2010 8:44 am
clock60 wrote:hi guys please share your ideas about this problem

The sum of first N-1 terms of AP is either zero or positive. but sum of first N terms is negative. What is the value of N?

A) Common difference is -4 and 7th term is last positive term
B) 25 is first term and there are 7 positive terms and all are integers
ao is B
Tn = a + (n-1) d
S(n-1) >=0 and S(n) < 0 => d is -ve
n = ?

Statement A :
d = -4
T(7) = a + 6d > 0 => a > -6d => a > 24
T(8) = a+7d <= 0 => a <= -7d => a <= 28

So, a can be one of (25,26,27,28), Therefor the value of n would not be same for all these cases.
--- Insufficient

Statement B :
a = 25
T(1), T(2)...,T(7) > 0

T(7) > 0
=> a + 6d > 0
=> 25 + 6d > 0
=> d > -25/6 => d = (-4,-3,-2,1) ------ 1

T(8) < 0 => a + 7d < 0
=> d < -25/7 = > d < -3.5 ------- 2

From (1) and (2), d = -4

Hence we have an AP where we know the first term and the common difference.
We can get an unique value for n. -- Sufficient
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by clock60 » Thu Dec 23, 2010 9:03 am
hi anshumishra
your solution it too smart for me
i stoped from the beggining
can you please elaborate or prove why d is -ve
S(n-1) >=0 and S(n) < 0 => d is -ve ,
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by anshumishra » Thu Dec 23, 2010 9:12 am
clock60 wrote:hi anshumishra
your solution it too smart for me
i stoped from the beggining
can you please elaborate or prove why d is -ve
S(n-1) >=0 and S(n) < 0 => d is -ve ,
Sure,
Although we never needed to use the conclusion (d > 0) in the statements given, here is how would you know ?

Since we know Sum of the first (n-1) terms were not negative, but when we added nth term to it, it became -ve. That means our AP is decreasing in value, hence the common difference is -ve.

Here is mathematical proof :

S(n-1) = (n-1)/2 [2a + (n-2) d] >= 0
=> 2a + (n-2)d > =0
=> 2a + (n-1) d - d > =0 ----------- (1 )

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

From (1) and (2), you can conclude that d is -ve.
Hope that helps.
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by clock60 » Thu Dec 23, 2010 10:01 am
got you 1st point but again question concerning..
So, a can be one of (25,26,27,28), Therefor the value of n would not be same for all these cases.

i understand that if
a=25 then a7=1 and a8=-3
a=26, then a7=2 and a8=-2, and so on till a=28
but i can`t understand what is at least the range of n-? why it is not 8
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by Anurag@Gurome » Thu Dec 23, 2010 10:26 am
clock60 wrote:The sum of first N-1 terms of AP is either zero or positive. but sum of first N terms is negative. What is the value of N?

A) Common difference is -4 and 7th term is last positive term
B) 25 is first term and there are 7 positive terms and all are integers
This solution is almost same as what anshumishra provided, but I tried to keep it more lucid and more correct.

Given: Sum of first (N - 1) terms, S(N - 1) of an AP is greater than equal to zero but sum of first N terms, S(N) is negative. Thus the AP must be a decreasing AP (thus common difference is negative) and N-th term is also negative. We have to find the value of N. Which done by solving these two inequalities, S(N - 1) ≥ 0 and S(N) < 0.

Say the first term be a and common difference be d.

Statement 1: Common difference is -4 and 7th term is last positive term.
Thus 7-th term is positive but 8-th term is non-positive and d = -4
=> (a + 6d) > 0 and (a + 7d) ≤ 0
=> (a - 24) > 0 and (a - 28) ≤ 0 ......... Replacing d = -4
=> a > 24 and a ≤ 28
=> 24 < a ≤ 28

Infinite number of values of a are possible and for each value of a there will be a solution to these inequalities S(N - 1) ≥ 0 and S(N) < 0. Thus more than one value of N is possible.

Statement 2: 25 is first term and there are 7 positive terms and all are integers.
Thus a = 25. As the AP is decreasing, the 7 positive terms must be the first 7 terms. And as the first 7 terms are integers, the common difference is an integer too.

Again 7-th term > 0 and 8-th term ≤ 0
=> (a + 6d) > 0 and (a + 7d) ≤ 0
=> (25 + 6d) > 0 and (25 + 7d) ≤ 0 ........ Replacing a = 25
=> d > -25/6 and d ≤ -25/7
=> -25/6 < d ≤ -25/7
=> -4.(something) < d ≤ -3.(something)

As d is an integer, only possible value of d is -4.
Now we have unique value of a and d. Thus we can uniquely solve the inequalities S(N - 1) ≥ 0 and S(N) < 0 for N.

Sufficient.

The correct answer is B.
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by anshumishra » Thu Dec 23, 2010 10:50 am
clock60 wrote:got you 1st point but again question concerning..
So, a can be one of (25,26,27,28), Therefor the value of n would not be same for all these cases.

i understand that if
a=25 then a7=1 and a8=-3
a=26, then a7=2 and a8=-2, and so on till a=28
but i can`t understand what is at least the range of n-? why it is not 8
I made a mistake to point where I assumed that there is only four possible values for a [25,26,27,28], but those are just the integral values of a, a could be anything between (24,28]). Anurag has rightly pointed that out.

However, to answer your problem :
Here are few possible sequences :
(25,21,17,13,9,5,1,-3,-7,-11,-15,-19,-23,-27,-31.....)
(26,22,18,14,...........................................................)
(27,23,19,15,...........................................................)

So, "N" here is the number of terms, when the sum of the first N terms become -ve.
like : 25+21+17+......+(-27).. and so on, when the sum becomes -ve.
Hopefully, you got the idea.
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by clock60 » Thu Dec 23, 2010 11:03 am
guys great thanks to both for wonderful solution and for patience, but for me the problem is not trivial
on real test i even will not attempt to solve it is too hard and long despite it is from GP software
just for curiosity
i got n-14 as a result (13,5<n<=14,5) does anybody solved it till the end to verify?
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by anshumishra » Thu Dec 23, 2010 2:04 pm
clock60 wrote:guys great thanks to both for wonderful solution and for patience, but for me the problem is not trivial
on real test i even will not attempt to solve it is too hard and long despite it is from GP software
just for curiosity
i got n-14 as a result (13,5<n<=14,5) does anybody solved it till the end to verify?
No Problem.

I didn't solve it till the end. That is the beauty with Data Sufficiency Questions.
If you are sure, you don't need to solve till end.

Here you may use this logic :
Consider two cases out of infinitely many :

(25,21,17,13,9,5,1,-3,-7,-11,-15,-19,-23,-27,-31.....)
(28,23,19,15,11,7,3,-1,-5,-9,-13,-17,-21,-25,-29,....)

Now Sum of an AP can be written as (we want it to be -ve for the n terms) :
n/2 * (First Term + nthTerm) < 0
=> nth Term < - (First Term)
From this you can conclude, that for the series starting with 25, the nth term will be less than what will it be for the series starting with 28.

{
Which means for the series starting with 25, I am looking for when the term becomes smaller than -25
Similarly for the series starting with 28, I am looking for when the term becomes smaller than -28
If both are decreasing by -4, it should be clear that for the 2nd series it will come later.
}

Think about this for a bit, if you still find you didn't get it, Let me know. I'll try to make it clearer.

Thanks
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