In an increasing sequence of 10 consecutive integers, the

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In an increasing sequence of 10 consecutive integers, the [#permalink]

07 Dec 2012, 04:31

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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

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Re: In an increasing sequence of 10 consecutive integers, the [#permalink]

06 Feb 2013, 03:05

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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]

07 Dec 2012, 04:38

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]

06 Feb 2013, 03:23

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.

Re: In an increasing sequence of 10 consecutive integers, the [#permalink] 06 Feb 2013, 03:23

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In an increasing sequence of 10 consecutive integers, the

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