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

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(#M40026447) TCS Numerical Ability question find sum Keep an EYE

solve
1-2+3-4+5-6........-98+99

Asked In TCS SHIVANG RANA (10 years ago)

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Read Solution (13) :

chaitanya  (10 years ago)

Group the terms as follows:
1 - 2 + 3 - 4 + 5 - 6 + 7 - ... + 97 - 98 + 99
= (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + ... + 98).

These are both arithmetic series.

Note that 1 + 3 + 5 + 7 + ... + 99 is an arithmetic series with a first term of 1 and a common difference of 2. It also has 50 terms. Since the sum of an arithmetic series, Sn, with a first term of a and a common difference of d is:
Sn = (n/2)[2a + (n - 1)d],

we see that:
1 + 3 + 5 + 7 + ... + 99 = (50/2)[2(1) + (50 - 1)(2)] = 2500.

2 + 4 + 6 + ... + 98 is also an arithmetic series (a = 2, d = 2, n = 49). Thus:
2 + 4 + 6 + ... + 98 = (49/2)[2(2) + (49 - 1)(2)] = 2450.

Therefore, the required sum is 2500 - 2450 = 50.

Is this solution Helpfull? Yes (20) | No (2)  

harshu  (10 years ago)

Ans: 50
(1+3+5......99)-(2+4+...........98)
from AP
1st series has 50 terms and second has 49 term
so
(50*(1+99)/2)-(49*(1+98)/2)
hence ans would be 50.

Is this solution Helpfull? Yes (11) | No (1)  

Chandan Kumar Nayak  (10 years ago)

In the whole expression u can see there are 50 terms[(1-2),(3-4)...(97-98) & 99]if we r taking each first 2 digits are a group. So for expression (1-2)+(3-4)+.....+(97-98)= -49 thus ans will be -49+99=50

Is this solution Helpfull? Yes (5) | No   

pritam banerjee  (10 years ago)

1 - 2 + 3 - 4 + 5 - 6 + 7 - ... + 97 - 98 + 99
= (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + ... + 98).

These are both arithmetic series.

Note that 1 + 3 + 5 + 7 + ... + 99 is an arithmetic series with a first term of 1 and a common difference of 2. It also has 50 terms. Since the sum of an arithmetic series, S_n, with a first term of a and a common difference of d is:
S_n = (n/2)[2a + (n - 1)d],

we see that:
1 + 3 + 5 + 7 + ... + 99 = (50/2)[2(1) + (50 - 1)(2)] = 2500.

2 + 4 + 6 + ... + 98 is also an arithmetic series (a = 2, d = 2, n = 49). Thus:
2 + 4 + 6 + ... + 98 = (49/2)[2(2) + (49 - 1)(2)] = 2450.

Therefore, the required sum is 2500 - 2450 = 50.

Is this solution Helpfull? Yes (4) | No (1)  

pandimurugan  (10 years ago)

odd series: n=50 ->n^2=50^2=2500;
even numbers=49->n(n+1)=49*50=2450;
odd-even=2500-2450=50;
answer:50

Is this solution Helpfull? Yes (3) | No   

Saurabh Mishra  (10 years ago)

ans:50
(1+3+5.....+99)-(2+4+......+98)=(2500)-(2450)=50
concept: sum of AP Series...

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

Rachana naik  (10 years ago)

ANS- 50
as per AP series..

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

Guru Pranav  (10 years ago)

(1+2+3+4+5+6.....+99) + (-4,-8,-12......-196)

4950-4900 = 50 Ans.

Is this solution Helpfull? Yes (2) | No   

dighe kajal m  (10 years ago)

50 as sum of 1st series=2500 using formua sn=n/2(2a+(n-1)d) and of 2nd series=2450
then diff bet them is 50

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Madhusmita  (10 years ago)

Answer is 50

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ARUN GUPTA  (10 years ago)

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

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

Arindam Dhara  (10 years ago)

(1+3+5+7.....99)-(2+4+6......98)
2500-2450
(sum of AP starting from 1 to 99 with diff 2)-(sum of AP starting from 2 to 98 with diff 2)
ans 50

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nadella deepthi  (7 years ago)

1-2+3-4+5-6...........-98+99
1-2=-1
3-4=-1
....
97-98=-1
98/2=49
-49+99=50
ans=50

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