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by harsh.champ » Fri Feb 19, 2010 12:13 pm
jagdeep wrote:If the sum of the consecutive integers from -35 to n inclusive is 150, what is the value of n?

A)
36

B)
39

C)
40

D)
60

E)
185
All the options choice are +ve so we can assume that n is +ve.

First of all we have to calculate the sum from -35 to 0.
Then,we can calculate the sum from 0 to n.

For negative sum,we get -[35 + 34 + 33 +......+ 3 + 2 + 1]
=-[(35)(35 + 1)]/2 = -630

For positive sum we have n(n+1)/2 =x.
Now,x -630 = 150
Hence,x=780
or n(n+1)/2 =780
=>n(n+1) = 1560 =39 x 40
Hence n=39 which is B.
Last edited by harsh.champ on Fri Feb 19, 2010 12:19 pm, edited 1 time in total.
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by abhi332 » Fri Feb 19, 2010 12:16 pm
If the sum of the consecutive integers from -35 to n inclusive is 150, what is the value of n?

A)
36

B)
39

C)
40

D)
60

E)
185

-35 to -1 will have 35 terms

0 ..... middle of -35 to +35

+35 to +1 will have 35 terms


also when we add above 2 sequence we get 0

-35 to +35 will contribute 0 sum and numbers of terms in sequence are 71 (-35 + -34 +.....-1+ 0 + 1+........34+35)

only choice E is above 71

IMO E is the ans
Last edited by abhi332 on Fri Feb 19, 2010 12:27 pm, edited 1 time in total.
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by harsh.champ » Fri Feb 19, 2010 12:24 pm
abhi332 wrote:
If the sum of the consecutive integers from -35 to n inclusive is 150, what is the value of n?

A)
36

B)
39

C)
40

D)
60

E)
185

-35 to -1 will have 35 numbers

0 ..... middle of -35 to +35

+35 to +1 will have 35 numbers


also when we add above 2 sequence we get 0

-35 to +35 will contribute 0 sum and numbers in sequence are 71 (-35 + -34 +.....-1+ 0 + 1+........34+35)

only choice E is above 71

IMO E is the ans
The answer would be B.
I think you made a mistake in understanding the problem.
numbers in sequence are 71 (-35 + -34 +.....-1+ 0 + 1+........34+35)
You have calculated the numbers in the sequence which sure is greater than 71(Actually it is 75(71+4) when n=39)
But in the question we are asked the value of n(which is a +ve integer)
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by abhi332 » Fri Feb 19, 2010 12:32 pm
Hello Harsh,

Thanks, I made a mistake writing terms as numbers, I have corrected that in the post.

Also, I have actually found the number of terms in original solution.

Can you tell me what I am missing
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by harsh.champ » Fri Feb 19, 2010 12:37 pm
abhi332 wrote:Hello Harsh,

Thanks, I made a mistake writing terms as numbers, I have corrected that in the post.

Also, I have actually found the number of terms.

Can you tell me what I am missing
Just note that sum of 36+37+38+39 = 150 which is the sum in the question.
Hence,the last term =n=39 which is the answer.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by abhi332 » Fri Feb 19, 2010 12:39 pm
Hey Harsh, Thank for your explanation,

I got my mistake :)
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by jagdeep » Fri Feb 19, 2010 12:57 pm
thanks guys....

Answer is B
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by shashank.ism » Sat Feb 20, 2010 7:50 am
jagdeep wrote:If the sum of the consecutive integers from -35 to n inclusive is 150, what is the value of n?

A) 36
B) 39
C) 40
D) 60
E) 185
Since -35 to 35 willl give the value = 0 and less than that will give value as -ve

so we have to check for values above 35
since 150 is a very small sum we can check it by addition also
36+37+38+39 = 150 (I applied this for fast calculation, hence this problem could be solved in 4 sec)
so Ans B
If there is a bigger no. we will go by AP process..
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by thephoenix » Sat Feb 20, 2010 8:17 am
jagdeep wrote:If the sum of the consecutive integers from -35 to n inclusive is 150, what is the value of n?

A)
36

B)
39

C)
40

D)
60

E)
185
clue1: 1st term=-35
clue2:no's r consecutive
implied clue summation of -35 to 35 will be 0 as all terms get cancel out
now summation of 36 to n is 150
now 36*5=180 >150
----> there can be only 4 no's
36+37+38+29=150 HENCE LAST TERM IS 39
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