Latest TCS Aptitude Question SOLUTION: What is the sum of all even integer between 99 and 301?

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(#M40033248) TCS Numerical Ability question Keep an EYE

What is the sum of all even integer between 99 and 301?

Asked In TCS Divya (10 years ago)

Solved RAKESH Read Solution (10)

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RAKESH  (10 years ago)

even integers betn 99 & 301 are starting from 100 & last is 300

Sum=100+102+104+...+300
its an AP with a=100,d=2,l=300,n=no. of terms
l=a+(n-1)*d => 300=100+(n-1)*2 => n=101

Sum = (n/2)*(a+l)= 101/2 *(100+300)= 101*200 = 20200

Is this solution Helpfull? Yes (37) | No (1)  

RAKESH  (10 years ago)

even integers betn 99 & 301 are starting from 100 & last is 300

Sum=100+102+104+...+300
its an AP with a=200,d=2,l=300,n=no. of terms
l=a+(n-1)*d => 300=100+(n-1)*2 => n=101

Sum = (n/2)*(a+l)= 101/2 *(100+300)= 101*200 = 20200

Is this solution Helpfull? Yes (5) | No   

Avinash  (10 years ago)

Ans: 20200
exp:
it is an ap Ist term is 100 last term =300
total no of terms={300-100}/2 + 1 =101 { +1 because both 100 and 300 are included }

sum of n terms = no of terms * { first term + last term}/2
=101*{100+300}/2
=101*400/2
=101*200
ans =20200

Is this solution Helpfull? Yes (2) | No   

Vivek Jain  (10 years ago)

series will be 100,102,104........300
so total number in the series =[(300-100)/2 + 1] because both the ends are inclusive n=101
sum of such series =n/2*[a+l]
=>101/2*[100+300]
=>101*200
=20200

Is this solution Helpfull? Yes (1) | No   

gaurav sningh  (10 years ago)

s = n[a+(n-1)*d/2]s

n = no of terms = 150
a = first term = 100
d = difference = 2

so the answer is 37350

Is this solution Helpfull? Yes (1) | No (3)  

suchithra  (10 years ago)

Answer: 20200
2(50+51+52........150)
2((150*151/2)-(49*50/2))
22650-245=20200

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Divya  (10 years ago)

is it 300=100+(n-1)*2 0r 300=100+(n-1)^2 ?

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Rakesh Kumar Kohli  (10 years ago)

Even integers between 99 and 301 are 100, 102, 104 ......,298,300
Their are 101 terms.
Their Sum =100+102+104+...+300
This an AP with first term a=100,common difference d =2 and last term l=300
Using formula Sum = (n/2)*(a+l)
We get Sum = (101/2)*(100+300)= 101*200 = 20200

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swetha  (10 years ago)

even integers between 99 and 301 starts from 100 and ends at 300.
now [(100+300)/2]*[[(300-100)]/2+1]
=200*101=20200

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koyel chatterjee  (10 years ago)

1st & last even integer between 99 & 301 are 100 & 300 respectively. so we can say that the nth term 300=100+(n-1)2(as nth term=a+(n-1)d, where a=1st term of series, n=no. of terms & d=common difference).
so from the above eqution we get n=101.
now sum of series=n/2[2a+(n-1)d].
sum of even integer between 99 and 301=101/2[2*100+(101-1)2]=20200

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