Latest TCS Aptitude Question SOLUTION: 1-2+3-4+5-6+...................-98+99=??

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1-2+3-4+5-6+...................-98+99=??

Asked In TCS kavitha (10 years ago)

Solved Subrata Guha Roy Read Solution (13)

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Read Solution (13) :

Sandeep kumar  (10 years ago)

ANS IS 50

taking + digits and - digits separately
for (1,3,5,7......97,99) and by applying AP WE GET SUM AS 2500
FOR (2,4,6....,96,98) AND BY APPLYING AP WE GET -2450
SO TAKING 2500-2450=50 ANS

Is this solution Helpfull? Yes (31) | No   

Triloki Shaw  (10 years ago)

1-2+3-4+5-6+........-98+99
= (1-2)+(3-4)+(5-6)+.....+(97-98)+99
= -1-1-1-1.............-1(49th)+99
= -49+99 (as every bracket gives -1 and total no. of bracket is 49 so -49)
= 50 ans.

Is this solution Helpfull? Yes (15) | No   

Abhisek Das  (10 years ago)

1-2+3-4+5-6+ . . -98+99
= (1+3+5+7...+99) - (2+4+6+8....+98)

According to AP, if we know the first and last term and the total numbers then,
AP = n/2(a+a(n)) where a is the first number and a(n) is the last number.

Total odd nos. between 1 to 99 = 50
Total even nos. between 2 to 98 = 49

=> (1+3+5+7...+99) - (2+4+6+8....+98) = {50/2(1+99) - 49/2(2+98)}
=> 25*100 - 50*49
=> 2500 - 2450
= 50

Answer is 50.

Is this solution Helpfull? Yes (13) | No   

Subrata Guha Roy  (10 years ago)

We have to divide this sum into two parts :


1+3+5+7......+99

And

+(-2)+(-4)+(-6)+(-8).......+(-98)

Using the A.P formula, we get,

L= a+(n-1)d

where L= Last term , a= First term , n= Number of terms , d= Difference

In first case,1+3+5+7......+99

a=1, L=99, d=2

So, 99=1+(n-1)2
=> (n-1)2 =98
=> n-1=49
=> n=50

So, the sum of 1+3+5+7......+99= n/2(a+L)=50/2(1+99)=25×100=2500
In the 2nd case,+(-2)+(-4)+(-6)+(-8).......+(-98)= -(2+4+6+8.....+98)

Here, a=2, d=2 and L=98

So, L=a+(n-1)d
=> 98=2+(n-1)2
=> (n-1)2=96
=> n-1=48
=> n=49

So the sum of (2+4+6+8.....+98)=n/2(a+L)=49/2(2+98)=(49×100)/2=49×50=2450

Therefore, 1-2+3-4+5-6+...................-98+99=(1+3+5+7......+99)-((2+4+6+8.....+98)=2500-2450=50

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Subbrata  (10 years ago)

(1+3+5+.....+99)-(2+4+6+........+98)
=50/2 (1+99)-49/2 (2+98)
=2500-2450
=50.........ri8...........

Is this solution Helpfull? Yes (3) | No   

Saurabh Nikam  (10 years ago)

taking 1 appart
(-2+3)+(-4+5)+..... ==> (1)+(1)+.....
1 to 99 means 99 digits..
neglecting 1 for a while
there are 98 digits...
98/2=49
49*1+1= 50
so, 50 is the ans

Is this solution Helpfull? Yes (2) | No   

Harish Marrapu  (10 years ago)

1st series::
1,3,5,........99
In A.P with d=2 and a=1
To find Sn find n?
As we know,An= a+(n-1)d
So,here 99=1+(n-1)2
2n-1=99 or 2n=100
n=50
Thus Sn=n/2(2a+(n-1)d)
so Sn=50/2(2+98)=25*100=2500
2nd series::
-2,-4,-6,.........-98
AP with d=2,a=-2 and n=49 from (An=a+(n-1)d)
So Sn=49/2(-4+(48*-2)=49/2(-100)=49*-50=-2450
HENCE Sum of both the series gives the ANSWER:::2500-2450=50

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Avanish kumar dubey  (10 years ago)

(1+3+........+99)-(2+4+.....+98)
apply A.P on both we get 2500-2450=50

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Guru Pranav  (10 years ago)

Answer is 50.

Is this solution Helpfull? Yes (2) | No (1)  

sangram badi  (10 years ago)

1-2+3-4+......97-98+99
-1+-1+.....+99= -49+99=50

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NURUL ANSAR  (10 years ago)

odd terms are +VE and even are -VE.
1+99=100
-2-98=-100
.
.
.
47+53=100
-48-52=-100
upto these all nullify 100-100=0
ans is=> 49+51-50=50

Is this solution Helpfull? Yes (1) | No   

tanmoy dasgupta  (10 years ago)

i think first 6 digit sum= -3. then -3+7=+4. similarly ..........+99=+50 ans.

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ashish kumar jindal  (10 years ago)

we can write this like,
(1+3+5+7........99)-(2+4+6+....98)
(2500)-(2450)-50

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