BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Algebra : computation with integer

Expert replies
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
Join the discussion
Source: — Problem Solving |

by GMATGuruNY » Sun Dec 02, 2012 4:01 am
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion

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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Join the discussion

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
Join the discussion