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A sequence consists of 16 consecutive even integers written [#permalink]

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14 Apr 2013, 11:54

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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]

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14 Apr 2013, 12:13

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
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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]

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14 Apr 2013, 12:50

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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
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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]

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14 Apr 2013, 13:30

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
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Re: A sequence consists of 16 consecutive even integers written [#permalink]

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15 Apr 2013, 02:58

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.
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Re: A sequence consists of 16 consecutive even integers written [#permalink]

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15 Apr 2013, 03:59

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.
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Last edited by Transcendentalist on 15 Apr 2013, 04:39, edited 1 time in total.

Re: A sequence consists of 16 consecutive even integers written [#permalink]

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15 Apr 2013, 04:26

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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
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Re: A sequence consists of 16 consecutive even integers written [#permalink]

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15 Apr 2013, 07:11

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]

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19 Jan 2015, 13:59

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Re: A sequence consists of 16 consecutive even integers written [#permalink]

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09 Mar 2016, 22:03

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Re: A sequence consists of 16 consecutive even integers written [#permalink]

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12 Sep 2016, 05:49

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

Another Method

Sum of 1st 8 Terms = 424 Average of those 8 Terms = \(\frac{424}{8} = 53\)

i.e. 4 Even Terms after 53 from the first part of series. Therefore, 4Th Term = 54 8th Term = 54 + 3 * 2 = 60

9th Term = 62 Sum from 9th till 16th Term = 62+64+66+68+... = 60*8 + 2+4+6+8+... +16 = 480 + 2(1+2+3+...+8) = 480 + 2* \(\frac{(8*9)}{2}\) (Sum of consecutive numbers from 1 to n = \(\frac{(n)*(n+1)}{2}\))

= 480+72 = 552 (C)
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Re: A sequence consists of 16 consecutive even integers written [#permalink]

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05 Oct 2016, 23:57

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

(a1+a8)/2*8=424 a1+a8=106 since nos. are consecutive integers a4+a5=a1+a8 No. between a4 and a5 is (a4+a5)/2=53 So a4=53-1=52 First term is 52=a1+(4-1)*2 a1=46 Eigtth term from last is 9th term from first So a9=46+(9-1)*2=62 and 16th term a16=46+(16-1)2 a16=76 So sum from ninth term to 16th term =(62+76)/2*8 =552

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