1+2+3+4+5+6+7+8+9+.......100.
Answers
Answered by
3
1+2+3+4+5+6+8+9+10.............100
By arthematic sum formula
n/2(2a+(n-1)d)
Where a is first term
d is common difference that is 1
And nis no of terms
100/2 {2(1)+(100-1)1}
50{2+99}
50(101)
5050
By arthematic sum formula
n/2(2a+(n-1)d)
Where a is first term
d is common difference that is 1
And nis no of terms
100/2 {2(1)+(100-1)1}
50{2+99}
50(101)
5050
Answered by
3
ANSWER:
The above series is the sum of first 100 natural numbers so it is given by:
==>100(100+1)/2
==>100(101)/2
==>5050.
The above series is the sum of first 100 natural numbers so it is given by:
==>100(100+1)/2
==>100(101)/2
==>5050.
Similar questions