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(#M40115955) TCS question Keep an EYE

.9 Find the sum of all the odd numbers starting from 1 and ending at the greatest 4 digit number.




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Asked In TCS kalyanikalyanam (10 years ago)

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RAHUL KUMAR  (10 years ago)

1+3+5+7+........+9999
from A.P. find tn.
so tn=5000
now Sn= n/2(a+l)
n=5000, a=1, and l= 9999
slove this ans is 25000000.

Is this solution Helpfull? Yes (10) | No (2)  

shaileygoel  (10 years ago)

Always remember that sum of first n odd number is always =n^2
Means n is to power of 2.
So ,in question the value of n is 9999(largest 4 digit number)
Therfore,9999^2=99980001

Is this solution Helpfull? Yes (6) | No (3)  

malav farkya  (10 years ago)

by using formula for the addition of odd numbers= ((l+1)/2)^2
l=9999(last term)
=((9999+1)/2)^2
=25000000 ans

Is this solution Helpfull? Yes (4) | No   

Ashok  (10 years ago)

we knw that greatest 4 digit is 9999
1+3+5+7+........+9999
from A.P. find tn.
so tn=9999
now Sn= n/2(a+l)
tn =a+(n-1)d
from this we get n=5000
therefore Sn=25000000

Is this solution Helpfull? Yes (2) | No   

JohnJeev  (10 years ago)

odd numbers starting from 1 and ending at the greatest 4 digit number : 1,3,5.....9999
Sum of first n odd numbers : n*n = 9999*9999 = 99,980,001

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