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
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!
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.
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
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