Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A sequence consists of 16 consecutive even integers written [#permalink]
14 Apr 2013, 10:54

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

68% (02:28) correct
32% (02:27) wrong based on 69 sessions

A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers?

Re: A sequence consists of 16 consecutive even integers written [#permalink]
14 Apr 2013, 11:13

atalpanditgmat wrote:

A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers?

488 540 552 568 584

provide the fastest way to solve this problem...........

the sequence can be expressed as: 2x, 2x+2,2x+4 ,... and generally if P=position(1,2,...) each term is 2x+2(p-1) The sum of the first 8 will be 8*2x+2(1+2+3+4+5+6+7)=8*2x+2*28=424 so 8x=396 and 2x=46 so the first number is 46. The sum of the last 8 will be from 9th place to 16th place so 46+2(9-1)+...+46+2(16-1)= 46*8+2(8+9+10+11+12+13+14+15)=46*8+92*2=552 C

PS:I used 2x as term because I wanted to be sure it was even Scroll down for my quick solution _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: A sequence consists of 16 consecutive even integers written [#permalink]
14 Apr 2013, 11:33

Expert's post

atalpanditgmat wrote:

A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers?

488 540 552 568 584

provide the fastest way to solve this problem...........

Let the first term be n

sequence for first 8 terms n, n+2, ................n+14

Re: A sequence consists of 16 consecutive even integers written [#permalink]
14 Apr 2013, 11:50

6

This post received KUDOS

Actually I have found a QUICK way, check this out. the first 8 can be written as: x , x+2,...,x+14 the second 8 can be written as x+16,x+18,...,x+30 As you notice, there is a gap of 16 between each number and its corrispondent 8 positions after 1st=x 9th=x+16 So if we add the sum of the gaps 16*8 to the sum of the numbers without the gap = 424 we have the result 16*8+424=552 _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: A sequence consists of 16 consecutive even integers written [#permalink]
14 Apr 2013, 12:01

Expert's post

Zarrolou wrote:

Actually I have found a QUICK way, check this out. the first 8 can be written as: x , x+2,...,x+14 the second 8 can be written as x+16,x+18,...,x+30 As you notice, there is a gap of 16 between each number and its corrispondent 8 positions after 1st=x 9th=x+16 So if we add the sum of the gaps 16*8 to the sum of the numbers without the gap = 424 we have the result 16*8+424=552

Thats it!

Actually i tried very hard to recall this, but i couldn't, as i learned it long back

Re: A sequence consists of 16 consecutive even integers written [#permalink]
14 Apr 2013, 12:30

The fastest way is to use the sum property of evenly spaced sets

(\frac{First Term + Last Term}{2}) * No of terms = Sum of the series

For the 1st eight set -(\frac{2n + 2n + 14}{2})*8 = 424. n = 23

For the 2nd eight set -(\frac{2n + 16 + 2n + 30}{2}) * 8 = 552 - which is the answer - C.

Remember, for this problem you can write out the sequence such as 2n, 2n+2, 2n+4, 2n+6 or you can use some formula - 2n + 2 (m-1).. to get the eight term in the sequence you will be 2n + 2*(8-1) = 2n + 14 and the 16 term in the sequence is 2n + 2*(16-1) = 2n + 30. For me the earlier part is quick and less error prone.. while the formula abstraction might be good for you.

//Kudos please, if this explanation is good _________________

Re: A sequence consists of 16 consecutive even integers written [#permalink]
15 Apr 2013, 01:58

Zarrolou wrote:

Actually I have found a QUICK way, check this out. the first 8 can be written as: x , x+2,...,x+14 the second 8 can be written as x+16,x+18,...,x+30 As you notice, there is a gap of 16 between each number and its corrispondent 8 positions after 1st=x 9th=x+16 So if we add the sum of the gaps 16*8 to the sum of the numbers without the gap = 424 we have the result 16*8+424=552

Thnaks Zarrolou, I am still lost somewhere. Am i missing anything basic mathematics in your explanation. Please simplify further. _________________

Do not forget to hit the Kudos button on your left if you find my post helpful.

Re: A sequence consists of 16 consecutive even integers written [#permalink]
15 Apr 2013, 02:59

Approach i used..

Sum of n even consecutive integers = n(n+1)

Sum of 8 consecutive even integers that start at some point after n = (n+8)(n+9)

Given => (n+8)(n+9) - n(n+1) = 424

16n +72 = 424 - General equation for sum of consecutive 8 digits where n is where counting starts

Solving we get n = 22 (Where the count starts)

Sum of 8 consecutive integers after n = 22 will be count starting at n1 = 22+8 = 30

So Answer = 16*30 +72 = 552

To break this down into a formula...

Sum of x consecutive even integers = 2xn + x(x+1)/2 (n = (First term of series/2) - 1)

Also for reference sakes

Sum of x consecutive odd integers = (x+n)^2 - (n)^2 (n=(first term of series - 1)/2)

I derived this so dunno if there is a better form of it. _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

Re: A sequence consists of 16 consecutive even integers written [#permalink]
15 Apr 2013, 03:26

1

This post received KUDOS

atalpanditgmat wrote:

Zarrolou wrote:

Actually I have found a QUICK way, check this out. the first 8 can be written as: x , x+2,...,x+14 the second 8 can be written as x+16,x+18,...,x+30 As you notice, there is a gap of 16 between each number and its corrispondent 8 positions after 1st=x 9th=x+16 So if we add the sum of the gaps 16*8 to the sum of the numbers without the gap = 424 we have the result 16*8+424=552

Thnaks Zarrolou, I am still lost somewhere. Am i missing anything basic mathematics in your explanation. Please simplify further.

Hi atalpanditgmat, let me explain my method.

We have 16 numbers even and CONSECUTIVE so they can be expressed as: x, x+2, x+4, x+6, x+8, x+10, x+12, x+14, x+16, x+18, x+20, x+22, x+24, x+26, x+28, x+30 Here they are. Now look at this: evrey number in the first 8, has a gap of 16 to its corrispondent 8 positions after. That's the trick. ie: first is x => 9 pos after => x+16 second is x+2 => 9 pos after =>x+2+16 So every number in the last 8 can be written as 1st+16, 2nd+16 and so on. The sum of 1st 2nd 3rd 4th ... 8th is 424, so the sum of 9th 10th ... 16 th or (using the trick) 1st+16, 2nd+16, 3rd+16, 4th+16, ... 8th+16 is 424(the sum of the numbers without 16)+16*8(the sum of 16s)

Let me know if it's clear now _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: A sequence consists of 16 consecutive even integers written [#permalink]
15 Apr 2013, 06:11

Expert's post

atalpanditgmat wrote:

A sequence consists of 16 consecutive even integers written in increasing order. The sum of the first 8 of these even integers is 424. What is the sum of the last 8 of the even integers?

A. 488 B. 540 C. 552 D. 568 E. 584

For any given AP : Sum of the n terms : n/2*(A1 + An) and An = A1+(n-1)d, where d = common difference,(2 in this case) Given : 8/2*(A1+A8) = 424 Now the sum of the last 8 terms = 8/2*(A9+A16)--> 4*(A1+8d+A1+15d) = 4*[(A1+A8)+16d] = 424+64*2 = 424+128 = 552