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# A sequence consists of 16 consecutive even integers written

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

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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
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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

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 [#permalink]  14 Apr 2013, 11:50
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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
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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

Zarrolou, thanks for reminding this. +1 by heart

Narenn
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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
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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.
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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.
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Last edited by Transcendentalist on 15 Apr 2013, 03:39, edited 1 time in total.
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Re: A sequence consists of 16 consecutive even integers written [#permalink]  15 Apr 2013, 03:26
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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]  15 Apr 2013, 03:35
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 [#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

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