Algebra : computation with integer

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Algebra : computation with integer

by meovang87 » Sun Dec 02, 2012 2:18 am
This question i take from OG 13 edition (number 172th on the book). I am looking for an alternative approach to the problem, other than the answer given in the book.

Question:
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 ?

Answer: 20,200
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by Brent@GMATPrepNow » Sun Dec 02, 2012 7:51 am
meovang87 wrote:TQuestion:
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 ?
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.

Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (since 150-50+1=101)

To evaluate 2(50+51+52+...+149+150) I'll add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101x200 = 20,200

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by Brent@GMATPrepNow » Sun Dec 02, 2012 7:53 am
meovang87 wrote: 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 ?
Alternatively, if we want to evaluate 2(50+51+52+...+149+150) (see above), we can evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200

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by ritind » Wed Dec 05, 2012 11:19 pm
There's a simple formula : s = (n/2)*(a+l)
where s = sum of numbers
n = number of terms
a = first term
l = last term
Looking at quest we know a=99 and l=301
No. of terms b/w 99 and 301 = 301-99 = 202
Out of 202 every alternate term is even
So n = 202/2 = 101
s = (101/2)*(99+301) = 20200