Latest TCS Aptitude Question SOLUTION: find the value of the expression 1-2+3-4+5-6+.......-96+97-98+99

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(#M40038142) TCS Numerical Ability question Keep an EYE

find the value of the expression
1-2+3-4+5-6+.......-96+97-98+99

Asked In TCS shashank ck (9 years ago)

Unsolved Read Solution (32)

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Read Solution (32) :

Avinash Gupta  (9 years ago)

(1+3+5...99)-(2+4+...98)
let it be n1-n2 where n1=1+3+5+7...99 & n2=2+4+6...+98
n1 & n2 are in A.P. so we will solve it seperately
In case of n1
...................
first term=1
last term =99
so,99=1+(n-1)2
n=50
so n1=sum=50(1+99)/2=2500
In case of n2
..................
first term=2
last term=98
so,98=2+(n-1)2
n=49
so sum=n2=49(2+98)/2=2450
so,n1-n2=2500-2450=50

Is this solution Helpfull? Yes (44) | No (1)  

Sarvesh Kumar Yadav  (9 years ago)

50
1-2+3-4+.....-96+97-98+99
total no are 99
-1-1-1(49 times)+99=50

Is this solution Helpfull? Yes (26) | No   

siva prasad  (9 years ago)

we know sum of first n odd numbers=n*n,first even numbers=n*(n+1),
then my answer is
1+3+.....+99=50*50
2+4+......+98=49(49+1)
now answer is=(50*50)-(49*50)=50.its simple

Is this solution Helpfull? Yes (12) | No (3)  

vikrant  (9 years ago)

we get 49 pairs of -1..like 1-2=-1 then 3-4=-1 so last 99-49=50

Is this solution Helpfull? Yes (9) | No (2)  

Anjali Kumari  (9 years ago)

(1+3+5+.....+99)-(2+4+6+....+98)
first we calculate the no of terms in first A.P. series
99=1+(n-1)*2
(99-1)/2=(n-1)
n=49+1=50
now we calculate the no of terms in second A.P. series
98=2+(n-1)*2
(98-2)/2=(n-1)
n=48+1=49
(1+3+5+...+99)-(2+4+6+...+98)
((n/2)*(a+l))-((n/2)*(a+l))
((50/2)*(1+99))-(49/2)*(2+98)
(25*100)-(50*49)
2500-2450=50

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

Rajat sadhu  (9 years ago)

50
(1+3+.....99)-(2+4+....98) both the series are in A.P.
using summation formula...
sum of odd terms=2500
sum of even terms=2450
ans=2500-2450=50

Is this solution Helpfull? Yes (2) | No   

Akash Ashok Mane  (9 years ago)

1+3+.....+99= 50*50(total no present in sum)^2;
2+4+.....+98=49*50(total no present in sum)*(total no present in sum+1)
=50*50-(49*50)
=50(50-49)
=50

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

aarthi  (9 years ago)

it is the sum of odd numbers-sum of even numbers
sum of odd numbers=n/2+n/2=(n^2)/4=(10000/4)=2500
sum of even numbers=n/2*(n/2+1)=50*51=2550
The difference is odd-even ...ans:2550-2500=50

Is this solution Helpfull? Yes (1) | No   

Abhishek nkumar  (9 years ago)

(1+3+5+......+.....+99)-(2+4+6+8+........98)
let us find the sum of AP of given series.
1+3+5+7+....99 2nd series =(2+4+6+....98);
=2450
99=1+(n-1)*2
n=50.
now sn = n/2*(a+l);
= 50/2*(1+99);
=2500;
now above eqn.
= 2500-2450
=50ans.

Is this solution Helpfull? Yes (1) | No   

Utsav Kumar  (9 years ago)

1-2=-1
3-4=-1
97-98=-1
so total we have 98/2=49 -1's
so
-49+99=50

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

ishita khare  (9 years ago)

1-2 gives -1
3-4 gives -1 and so on
therefore we have -1-1-1.........49times+99= -49+99=50

Is this solution Helpfull? Yes (1) | No   

susmita dutta chowdhury  (9 years ago)

99 as it is an odd number so (99+1)/2=50 ans.

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

rahul  (9 years ago)

from 1 to 98 there are 98 numbers.
so 98/2 = 49
1-2+3-4+5-6+.......-96+97-98+99
= -49 + 99
=50

Is this solution Helpfull? Yes (1) | No   

tirupathi reddy  (9 years ago)

50
The given expression can be written as
= (1+3+5+....+99)-(2+4+6+....+98)

= 50/2(100) - 49/2(100)

= 50(50-49)

= 50

Is this solution Helpfull? Yes | No (1)  

lakshman singh  (9 years ago)

answer=50;
sum of all odd term from 1 to 99 - sum of all even term 2 to 98;
apply simple A.P. series to get sum ,TOTAL ODD TERM ARE 50 &EVEN TERM ARE 49;
=50*50-50*49=50;

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BISWAJIT NATH  (9 years ago)

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

Is this solution Helpfull? Yes | No   

yatin sharma  (9 years ago)

for odd numbers : 1+99 =100 ; 3+97 =100 ; 5+95=100 ...so on
for even numbers : -2 -98 = -100 ; -4 - 96 = -100 ; -6 -94 = -100 ...so on
hence in the middle there will be only 49 + 51 =100 and -50 implies 100-50 = 50

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bhavana  (9 years ago)

sum of odd numbers is 2500
sum of even numbers is 2450
2500-2450=50

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deepak kumar  (9 years ago)

1-2+3-4+.........-98+99
1+3+5......._(2+4+.....98)
2500-2450=50

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Zishan Khan  (9 years ago)

make two series of A.P by this expression
such that
1+3+5.....+97+99-(2+4+6....+96+98)
50/2(2*1+(50-1)2) - 49/2(2*2+(49-1)2) = 50

Is this solution Helpfull? Yes | No   

MUBEENA.J  (9 years ago)

For every two value,we have got -1.for example,1-2=-1,3-4=-1, so we have got
-49+99=50

Is this solution Helpfull? Yes | No   

thirumagal  (9 years ago)

(1+3+5+.....+99)-(2+4+6+....+98)
n A.P. series
a=1 and l=99
99=1+(n-1)*2
where n=50
(99-1)/2=(n-1)
n=49+1=50
now we calculate the no of terms in second A.P. series
98=2+(n-1)*2
(98-2)/2=(n-1)
n=48+1=49
(1+3+5+...+99)-(2+4+6+...+98)
((n/2)*(a+l))-((n/2)*(a+l))
((50/2)*(1+99))-(49/2)*(2+98)
(25*100)-(50*49)
2500-2450=50

Is this solution Helpfull? Yes | No   

Ashish Kumar  (9 years ago)

Up to 98 there are 98/2 times sum of -1. so sum = -49.
Last term is 99
so value of expression = -49 + 99 = 50

Is this solution Helpfull? Yes | No   

Jit Banerjee  (9 years ago)

1-2+3-4+5-6+.......-96+97-98+99
=(1-2)+(3-4)+(5-6)+.....+99
=-1-1-1.......+99
now for 1st 6 terms there are 3 no. of -1's
so for 98 terms there are 98/2=49 no. of -1's
+99(last term)
=-49+99
=50(Ans)

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Madhurima Chakraborty  (9 years ago)

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

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TANVI  (9 years ago)

1+3+.....+99=50*50
2+4+......+98=49(49+1)
now answer is=(50*50)-(49*50)=50.

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Dharmendra Shukla  (9 years ago)

sum of 50 terms-sum of 49 terms=2500-2450=50

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Pooja  (9 years ago)

answer is 0.

Is this solution Helpfull? Yes | No (4)  

Bedouin paul  (9 years ago)

There are 49 sets which results to 1...(-2+3)+(-4+5)+....+(-98+99)+1=49+1=50

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shilpi saini  (9 years ago)

0 ... Its an a.p of odd minus even terms

Is this solution Helpfull? Yes | No (2)  

Bhargav  (9 years ago)

0
Taking odd numbers common and even numbers common then expression becomes
(1+3+5...99)-(2+4+6+98)
Two sequences are in AP.
So sum of terms in AP (a+l)/2 if first term is a and last term is l.
Now applying formula
(1+99)/2-(2+98)/2=0

Is this solution Helpfull? Yes | No (1)  

Pooja  (9 years ago)

1-2=-1
3-4=-1
so,from 1 to 98 -49
so,99-49=50
ans=50

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