Sum of first 50 even numbers is..

[Guest]

The sum of the first 50 positive even integers is 2550 .What is the sum of even numbers from 102 to 200 inclusive?

[editor: the problem was originally transcribed incorrectly; i have fixed it.]

1)5100
2)7550
3)10100
4)15500
5)20100

Please also suggest on with below question :

The sum of first 50 even numbers is 650 [using Formula :m/2 (m/2+1)] ,why is it 2550 ?

[San]

The sum of first 50 even numbers is 2550 .What is the sum of even numbers from 102 to 200 inclusive?

1)5100
2)7550
3)10100
4)15500
5)20100

Please also suggest on with below question :

The sum of first 50 even numbers is 650 [using Formula :m/2 (m/2+1)] ,why is it 2550 ?

The answer is 7550.
"The sum of first 50 even numbers is 2550" ---- means that 1 to 100 numbers, there are 50 even numbers (ex; 2,4,6,8,10,......92,94,96,98,100). if you add 2+4+6+8+........98+100 (the even numbers), the answer will be 2550. The second part: "What is the sum of even numbers from 102 to 200 inclusive?" we knew that for first 50 even number is 2550, then think about that: 102 is 100 more than 2, 104 is 100 more than 4, 106 is 100 more than 6,... 198 is 100 more than 98, and 200 is 100 more than 100. so, 102+104+106+108+110=530 which is 500 more than 30 (2+4+6...+10), 112+114+.....+120 it will also 500 more than 80 (12+14+....+20). Thus, 122+124+....130 will also be 500 more than 22+24+....+30, then you add those 500 all together (in fact there are 10 of 500 number, start from 102+....110, 112+....+120, 122+...+130,......190+....+200), you will have 5000.

Remember the sum of first 50 even numbers is 2550 than you add 5000 to it , now the winner is 7550 :)

[Guest]

thank you!

[RonPurewal]

Please also suggest on with below question :

The sum of first 50 even numbers is 650 [using Formula :m/2 (m/2+1)] ,why is it 2550 ?

that's the first time i've ever seen a formula for that.

you aren't using the formula correctly; the way it's written, "m" must stand for the greatest number out of your list of even numbers, NOT the number of terms in the sequence.
therefore, "m" is 100, and so your formula gives (50)(50 + 1) = 2550, as expected.

[RonPurewal]

here's a compact presentation:

102 + 104 + 106 + ... + 200
=
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 100)
=
(100 + 100 + 100 + ... + 100) + (2 + 4 + 6 + ... + 100)
=
(fifty 100's) + (2550 as given)
=
5000 + 2550
=
7550

[imanemekouar]

RON 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

[agha79]

Another approach

Since these are consecutive even numbers from 102 to 200 inclusive we can find the middle number . In this case it will be sum of (150 + 152) / 2 = 151. Now we can multiply 151 x 50 and get 7550. The number 50 is total number of integers between 102 to 200 inclusive.

I hope this helps

[robert.edwards.jr]

RON 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

This is close. But to find the number of terms you need to add 1 to your result. The formula should be:

# of Terms = ([First - Last] / 2) + 1

[esledge]

It seems we have two threads going on the same question. Here's the other.

the-sum-of-first-50-even-numbers-is-2550-what-is-the-sum-of-t9317.html

[andy_11_30]

here's a compact presentation:

102 + 104 + 106 + ... + 200
=
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 100)
=
(100 + 100 + 100 + ... + 100) + (2 + 4 + 6 + ... + 100)
=
(fifty 100's) + (2550 as given)
=
5000 + 2550
=
7550

Ron,


why do we require first part of the question ?

- > 102 to 200 , total terms are 50 in number
- > Since A.P [ (200+102)/2 ] = 151 = Average
- > Avg = Sum / n
- > Sum = 151 * 50 = 7550 !!

Simple...!!

[rockrock]

I was thinking the same thing? Is the first part just there to confuse you? As long as you have first and last number, the number of integers and the increment (if different than 1), you can solve for the sum.... seems like it was perfectly defined regardless of the other info.

[adiagr]

here's a compact presentation:

102 + 104 + 106 + ... + 200
=
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 100)
=
(100 + 100 + 100 + ... + 100) + (2 + 4 + 6 + ... + 100)
=
(fifty 100's) + (2550 as given)
=
5000 + 2550
=
7550

Ron,


why do we require first part of the question ?

- > 102 to 200 , total terms are 50 in number
- > Since A.P [ (200+102)/2 ] = 151 = Average
- > Avg = Sum / n
- > Sum = 151 * 50 = 7550 !!

Simple...!!

Andy, You are right. using AP formula, question can be done without 1st part.

[RonPurewal]

I was thinking the same thing? Is the first part just there to confuse you? As long as you have first and last number, the number of integers and the increment (if different than 1), you can solve for the sum.... seems like it was perfectly defined regardless of the other info.

the gmat is generally not going to contain problems that significantly advantage people who have memorized common formulas.

it's possible that there will be problems that you can solve more or less immediately if you have memorized obscure formulas (e.g., GCF of x&y times LCM of x&y = x times y), but, don't forget, one of the principal purposes of this test is to contain problems for which memorization poses no advantage (and, in many cases, actually a disadvantage, by testing problems in which unusual restrictions or exceptions render the memorized rules incorrect).

therefore, the initial statement is presumably here as a way to level the playing field in favor of people who have not memorized the aforementioned formula for arithmetic progressions.

by the way, is this really an official problem?
it doesn't seem like one, but the inclusion of the first statement makes it seem much more like an authentic problem. (the test would certainly not contain a problem like this one WITHOUT the first statement, since that would advantage people who have memorized the formulas.)

[rockrock]

thanks that was a helpful response! it does help to think of it without the formula as well, since that could come in handy when it comes to word translations problems where you have to come up with the relationship yourself.

[mschwrtz]

Glad Ron's response was helpful.