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1-2+3-.................-100 find the value
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- 1-2+3-4.....+99-100 ,this is the sequence
now, we divide the series into 2 groups..
(1+3+5+7...+99)-(2+4+6+8...+100)
using A.P sum formula,
(1+3+5...+99)=50/2(1+99) [s(n)=n/2(a+l)]
=2500
again for another sequence
(2+4+6...+100)=50/2(2+100)
=2550
the original sequence was=(1+3+5+7...+99)-(2+4+6...+100)
= 2500-2550
= -50 (ans). - 11 years agoHelpfull: Yes(22) No(0)
- 1-2=-1
3-4=-1..... similarly -1-1.....50times=-50 - 11 years agoHelpfull: Yes(9) No(1)
- divide the sequence into 2 groups
sum of odd numbers =n*n (n=number of odd numbers)=2500
sum of even numbers =(n+1)*n (n=number of even numbers)=2550
ans is 2500-2550=-50 - 10 years agoHelpfull: Yes(2) No(0)
- it will be-50
- 11 years agoHelpfull: Yes(1) No(0)
- -50
- 11 years agoHelpfull: Yes(1) No(0)
- -50 ans...
- 11 years agoHelpfull: Yes(0) No(0)
- ans=-50
...50*(-1) - 11 years agoHelpfull: Yes(0) No(0)
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