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In an increasing sequence of 10 consecutive integers, the

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In an increasing sequence of 10 consecutive integers, the [#permalink] New post 07 Dec 2012, 03:31
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In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?

(A) 585
(B) 580
(C) 575
(D) 570
(E) 565
[Reveal] Spoiler: OA
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 07 Dec 2012, 03:38
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Walkabout wrote:
In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?

(A) 585
(B) 580
(C) 575
(D) 570
(E) 565


Say our 10 consecutive integers are x, x+1, x+2, ..., x+9.

x+(x+1)+(x+2)+(x+3)+(x+4)=560.
(x+5)+(x+6)+(x+7)+(x+8)+(x+9)=?

Notice that the 6th term is 5 more than the 1st term, the 7th term is 5 more than the 2nd term, ..., the 10th term is 5 more than the 5th term, thus the sum of the last 5 terms is 5*5=25 greater than the sum of the first 5 terms. Therefore the answer is 560+25=585.

OR:
x+(x+1)+(x+2)+(x+3)+(x+4)=5x+10=560.
(x+5)+(x+6)+(x+7)+(x+8)+(x+9)=5x+35=(5x+10)+25=560+25=585.

Answer: A.
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 06 Feb 2013, 02:05
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each of 5 last number is greater than each of 5 first number by 5.

we have 5 times 5=25
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 06 Feb 2013, 02:23
Walkabout wrote:
In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 560. What is the sum of the last 5 integers in the sequence?

(A) 585
(B) 580
(C) 575
(D) 570
(E) 565




x+(x+1)+(x+2)+(x+3)+(x+4)=5x+10=560.
(x+5)+(x+6)+(x+7)+(x+8)+(x+9)=5x+35=(5x+10)+25=560+25=585.
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 25 Mar 2014, 00:26
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Let the 5 consecutive numbers be
x , x+1 , x+2 , x+3 , x+4

Given that there sum = 560

5 (x+2) = 560 ............ As they are consecutive; so sum will be middle term * 5

x+2 = 112

x = 110

So next five nos will be

115 , 116 , 117 , 118 , 119

There sum = 117 * 5 = 585 = Answer = A
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 28 May 2014, 08:36
tanviet wrote:
each of 5 last number is greater than each of 5 first number by 5.

we have 5 times 5=25


pure genius 10 sec approach! :)
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 28 May 2014, 08:56
560 + 5*5 = 585, Option A)
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink] New post 20 Jul 2014, 19:26
Average of 1st five numbers:

560/5 = 112 Since there are 5 numbers, the average is in the middle of the set, thus being the 3rd number.

110 111 112 113 114 ... now you just add the other consecutive 5 to equal 585.

This is a more organic approach if algebra is rusty.
Re: In an increasing sequence of 10 consecutive integers, the   [#permalink] 20 Jul 2014, 19:26
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