The sum of first 50 even numbers is 2550 .What is the sum of

[imanemekouar]

pLease correct me if I'm wrong:

Please correct if i m wrong:


Sum=average*numbers of terms

Numbers of term=(200-102)/2 =43

Average= (FIRST +LAST)/2 = (200+102)/2 = 151

SUM = 151*43 = 6493

[esledge]

Please post the full text of the question. Is it "what is the sum of the next 50 even numbers?" And what are the choices?

Sum=average*numbers of terms

Numbers of term=(200-102)/2 =43

Your error is here. The number of even numbers between 102 and 200, inclusive, is
(200-102)/2 + 1 = 98/2 + 1 = 49 + 1 = 50.
{You divide the range by 2 since only half of any set of consecutives is even, and "add one before your done" to include 102 as one of the terms you are counting.}

Also, think about this: There are 50 positive even numbers up to and including 100 (i.e. 2, 4, 6, ...., 98, 100). The set from 102 to 200 is just the first set shifted up by 100.

Thus, Sum = Average * Number of Terms = 151 * 50 = 7550. Check it another way: if each term is 100 more than the corresponding term in the 2 to 100 set, and there are 50 terms, this sum should be 5000 greater than the 2550 sum given.

[praffulpatel]

Hi,

Sorry to open this topic again.

Full question is here.

1. The sum of first 50 even numbers is 2550. What is the sum of even numbers from 102 to 200 inclusive?
A. 5100
B. 7550
C.10100
D.15500
E. 20100

I reached the answer to this question as described below.

Sum of first n even natural numbers = n(n + 1)

so the first 50 integer's sum = 50(51) = 2550 as given in the question stem.

Sum of first 100 integer (2550 covers integers till 100, and 102 and 200 are inclusive) = 100(101) = 10100

Sum of integers from 102 to 200 therefore = 10100-2550 = 7550.

Can someone confirm if my approach is correct?

[RonPurewal]

here's a thread with approximately a million posts on this problem:

sum-of-first-50-even-numbers-is-t4166.html

if you have any more questions after reading that thread, post them over there, not here. this thread is now locked, so that we don't have tons of different threads on the same problem.