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A sequence consists of 16 consecutive even integers written
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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? A. 488 B. 540 C. 552 D. 568 E. 584
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Re: A sequence consists of 16 consecutive even integers written
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14 Apr 2013, 12:50
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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Re: A sequence consists of 16 consecutive even integers written
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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 \(2x+2(p1)\) 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 \(46+2(91)+...+46+2(161)= 46*8+2(8+9+10+11+12+13+14+15)=46*8+92*2=552\) 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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Re: A sequence consists of 16 consecutive even integers written
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14 Apr 2013, 12:33
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 Sum = 424 = 8n + (0 + 2 + 4 + 6 + 8 + 10 + 12 + 14) > 424 = 8n + 56 > 8n = 368 > n=46 So the first term is 46 last 8 terms will be n + 16 .........................n + 30 i.e. 62, 64, ..................,76 This is an Arithmatic Progression with first term = n + 16 last term = n + 30 number of terms = 8 Sum = \(\frac{number of terms(first term + last term)}{2}\) Sum = \(\frac{8(62 + 76)}{2}\) Sum = \(\frac{8(138)}{2}\) Sum = 138 X 4 = 552 Regards, Narenn
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Re: A sequence consists of 16 consecutive even integers written
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14 Apr 2013, 13:01
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 Zarrolou, thanks for reminding this. +1 by heart Narenn
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Re: A sequence consists of 16 consecutive even integers written
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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 (m1).. to get the eight term in the sequence you will be 2n + 2*(81) = 2n + 14 and the 16 term in the sequence is 2n + 2*(161) = 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
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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
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Updated on: 15 Apr 2013, 04:39
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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Re: A sequence consists of 16 consecutive even integers written
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15 Apr 2013, 04:26
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
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15 Apr 2013, 04:35
Aha! I get the right path now. Thanks thanks a lot....
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Re: A sequence consists of 16 consecutive even integers written
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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+(n1)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 C.
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Re: A sequence consists of 16 consecutive even integers written
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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
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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=531=52 First term is 52=a1+(41)*2 a1=46 Eigtth term from last is 9th term from first So a9=46+(91)*2=62 and 16th term a16=46+(161)2 a16=76 So sum from ninth term to 16th term =(62+76)/2*8 =552



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Re: A sequence consists of 16 consecutive even integers written
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03 Mar 2017, 11:16
The answer is C first 8: 424 last 8: 424 + 8 * 16 = 552



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