For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2. What is the sum of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
Source: OG-12
[spoiler]OA: (B) [/spoiler]
Integers..
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Hi Elena,
The problem can be solved using different methods. Here are a couple of them.
nth term in an Arithmetic Progression = First term +(n-1)*(Common difference)
300 = 100 + (n-1)*2 => n = 101
Sum of n terms = (n/2)*(first term + last term) = (101/2)*(100+300) = 20200
= 2*(50+51+52+...150)
= 2*((1+2+3+4+...150)-(1+2+3+.....49))
= 2*((0.5*150*151)-(0.5*49*50))
= ((150*151)-(49*50))
= (50*((151*3)-49))
= 50*404
= 20200
The problem can be solved using different methods. Here are a couple of them.
Consecutive even integers are in Arithmetic Progression, where the common difference of the series is 2, First term 100 and Last term 300.Method 1)
nth term in an Arithmetic Progression = First term +(n-1)*(Common difference)
300 = 100 + (n-1)*2 => n = 101
Sum of n terms = (n/2)*(first term + last term) = (101/2)*(100+300) = 20200
sum of all the even integers between 99 and 301 = 100+102+104+....300Method 2)
= 2*(50+51+52+...150)
= 2*((1+2+3+4+...150)-(1+2+3+.....49))
= 2*((0.5*150*151)-(0.5*49*50))
= ((150*151)-(49*50))
= (50*((151*3)-49))
= 50*404
= 20200
Anil Gandham
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Use the AP formula directly for sum with a1 = 100,an = 300 and d =2.
With this info find n and the Sn = n/2 * (a1 + an).
With this info find n and the Sn = n/2 * (a1 + an).