Sum of even integers

[tester2]

The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200 inclusive?

a)5100 b) 7550 c) 10100 d) 15500 e) 20100

 

 

OK, sometimes these ridiculous number sums come up and i never know how to hand them. I guessed the correct answer of B only because I figured A was a trap. How do you do this problem?

Thanks!

[Shooter]

I am not sure how to use the given first information (i.e) 2550. I did as follows:

 

number of terms b/w 102 and 200 (200-102+1) = 98. 50 terms are even

n=50;a=102;d=2

 

substitute in n/2[2a+(n-1)d]

Gives you 7550

[shobby]

(1) For 1 to 100, terms are 2, 4, 6, ..so on

(2) For 101 to 200, terms are 102, 104, 106, .. so on

Every term in (2) is 100 more than its corresponding term in (1). So, sum of terms of (2) is 2550 + 50(100) = 7550

[Retake_GMAT]

Just apply formula for AP.No tricks at all!!

200 = 102 + (n-1) d

200= 102 + 2n -2

202=102 + 2n

100=2n

n = 50

25 * ( 2*102 + 49*2)= 25* ( 204 + 98) = 302*25 =7550

[ankitharish]

The first info can be used to derive the formulae for AP.

But better learn the formulae plain and simple.

[Gymat]

What is AP? I'm not sure that I understand where did you get D=2?

[bose]

sum of first 50 even # i.e. from 2-100=2550

sum of 102 - 200=here each term is added with 100, so 100*50+2550

since 50 even terms are there in 102-200

ans 7550

[gmat_magic_prep]

sum of first 50 even # i.e. from 2-100=2550

sum of 102 - 200=here each term is added with 100, so 100*50+2550

since 50 even terms are there in 102-200

ans 7550

sum of first 50 +ve even integers = 2550

2 + 4 + .. + 100 = 2550

Given sum = G = 2(1+2+ .. + 50) = 2550

 

Required sum = S = 102 + 104 + .. + 200

=> S = 2(51 + 52 + ... + 100)

 

Now, G + S = 2550 +S = 2(1+ .. +50) +2(51+ .. +100)

=> S + 2550 = 2(1+2+3+ ..100)

= 2 * 100 * 101 / 2 = 10100

=> S = 10100 - 2550 = 7550

 

This appears long but the easiest way.

[mail2jkd]

Step 1: Find the average of the even series - (102+200)/2 = 151

 

Step 2: Find the number of even integers in the series - Easy way, 200-102=98+1=99 integers in the series. 99/2+1=50 even integers because it starts and ends with a even number.

 

50x151=7550. Ans:B

 

Or you can use the AP nth digit formula to get the number of even integers in a series.

 

Tn=(a+n-1)d

200=102+(n-1)2

n=50

[gcvbt7t]

7550 is the an swer

[karthikvela]

Yeah the answer is 7550. The simplest way is given by mail2jkd. The same kind of problem can be found in OG11 PS portion Q.no:204.