Latest TCS Aptitude Question SOLUTION: sum of all even numbers between 99 to 301 is?

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(#M40019745) TCS Numerical Ability question Keep an EYE

sum of all even numbers between 99 to 301 is?

Asked In TCS lokeshwari (10 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

Aditya  (10 years ago)

we have terms :100,102.....300
tn=300,a=100,d=2
tn=a+(n-1)d
tn=101
Sum=n/2(a+l)=101/2 *(100+300)
=20200

Is this solution Helpfull? Yes (36) | No (1)  

k.m  (10 years ago)

i think answer is 20200 bcoz there is total 101 even terms in between 99 & 301.
so by applying formula s(n)=n/2*(a+l) i.e 101/2(100+300)=20200 where a & l are 1st and last term.
so answer is 20200..

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

ABHISHEK KUMAR MISHRA  (10 years ago)

Here no are: 100+102+104+.......+300
The above sequence is in A.P where common difference(d) = 2
first term(a) = 100
total no of terms = 101
so sum of sequence is n/2*(2a + (n-1)*d) = 101/2*(200+(101-1)*2) = 20200

Is this solution Helpfull? Yes (5) | No   

manoj chauhan  (10 years ago)

even terms from the given numbers will be 100,102,104.........,300
clearly all terms are in AP whose first term is a=100 and common difference d=2, last term is l=300
we know that l=a+(n-1)d
300=100+(n-1)2 solving this we will get a=101
sum of nth term is Sn=n/2(a+l)
Sn=101/2(100+300)
20200 this is answer

Is this solution Helpfull? Yes (2) | No   

bhanu  (10 years ago)

we have AP 100,102,....,300
a=100,d=2,tn=300
so,n=101(apply formula-tn=a+(n-1)d) and hence
sn=n/2*(a+tn)
so,sum=20200

Is this solution Helpfull? Yes (1) | No   

sujeet kumar keshri  (10 years ago)

100,102,...................300
let no of even term =n
nth term=100+(n-1)2=300
n=99
sum=(99/2){2*100+(99-1)2=19602 ans

Is this solution Helpfull? Yes | No (9)  

CHETAN BHANGALE  (10 years ago)

i think 20200.

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muthu narayanan  (10 years ago)

100+102......+300...a=100,l=300,n=51..using n/2(a+L) this formulae,ans:20200

Is this solution Helpfull? Yes | No (1)  

Dipak Liladhar Patil  (10 years ago)

nth term = a+(n-1)d
where a=first term d=difference
hence
300 = 100+(n-1)*2
after solving, n=101
sum = n/2(2*a+(n-1)*d)

sum=101/2(2*100+(101-1)*2)

sum = 20200.

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richa ahuja  (10 years ago)

series will b
100 102 104 106....300
nw dis is the series with frst term 100 and last term 300 with common difference d= 2
let no of terms b n
thus,300=100+(n-1)d
d=2
n=101
sn=n/2(2a+(n-1)d)
a=100
sn=20200

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