What is the sum of the first 250 even integers?

Answer 1

The formula for finding the sum (Sn) of a certain number of terms in an arithmetic is given here:

Sn = n(a1 + an) / 2

n = number of terms

a1 = the first term

an = the nth term

But there is a catch. (There is always a catch.) We need to know what the last term (an) in the sequence is. If we were adding the first 100 numbers, the last term in the sequence would be 100. We can see that; we can determine it by simple inspection. But we've been asked to add the first 250 even numbers. Further, what if we were asked to find the last term in the series of the first 200 7th numbers (7, 14, 21, etc.) so we could work that problem? We'd better have a way to find the last term of a sequence without a struggle. There is a handy formula to do that, and it's here:

an = a1 + (n-1)d

a1 = the first term

an = the nth term

n = number of terms

d = the common difference

Let's work that remembering that for our sequence of the first 250 even numbers, the common difference is 2 (because even numbers have a common difference of 2).

an = 2 + (250-1)2 = 2 + (249)2 = 2 + 498 = 500

You might have "seen" that the last term in the sequence was 500, but try the formula for finding that last term in a series with a common difference that is "less easy" than the one here. The "easy" questions are probably not going to be on the test anyway. The "harder" ones will probably show up. They usually do. Let's move on and find our answer.

We've got our last term, so we can plug that into the equation to crank out an answer:

Sn = 250(2 + 500) / 2 = 250(502) / 2 = (125500) / 2 = 62750

Try a couple more as soon as you can to "lock in" the methods described so you'll be ready for the test.

Another way is to notice that this particular question is asking for twice the sum of the first 250 integers, The standard formula for the sum of the first n integers (using Sn, the first formula above) is n(n+1)/2 ) so doubling gets rid of the 1/2 and you get 250 x 251 =62750