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

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Question Stats:

71% (02:55) correct
29% (02:36) wrong based on 114 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

Re: A sequence consists of 16 consecutive even integers written [#permalink]
19 Jan 2015, 12:59

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