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# In an increasing sequence of 10 consecutive integers, the

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

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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]

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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]

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07 Dec 2012, 03:38
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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

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.

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

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06 Feb 2013, 02:23
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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

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]

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20 Jul 2014, 19:26
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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.
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink]

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19 Feb 2015, 14:27
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This is my approach:
X+x+1+x+2+x+3+x+4=560
5x+10=560
x=110
Total sum=10/2(2(110)+1(10-1))=5(229)=1145
Total sum- sum of first five terms=1145-560=585
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink]

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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]

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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]

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02 Feb 2016, 06:55
thangvietnam
deepakpandey028

thx so much for ur incredible solution, cud i ask u to give a broader explanation?

thangvietnam wrote:
each of 5 last number is greater than each of 5 first number by 5.
we have 5 times 5=25

deepakpandey028 wrote:
560 + 5*5 = 585, Option A)
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In an increasing sequence of 10 consecutive integers, the [#permalink]

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02 Feb 2016, 07:04
studentsensual wrote:
thangvietnam
deepakpandey028

thx so much for ur incredible solution, cud i ask u to give a broader explanation?

thangvietnam wrote:
each of 5 last number is greater than each of 5 first number by 5.
we have 5 times 5=25

deepakpandey028 wrote:
560 + 5*5 = 585, Option A)

Assume that the terms of the sequence are

x
x+1
x+2
x+3
x+4

x+5
x+6
x+7
x+8
x+9

When you compare x with x+5 or x+1 with x+6, the difference is always = x+5-x = 5

Thus you can make 5 pairs
(x,x+5)
(x+1,x+6)
(x+2,x+7)
(x+3,x+8)
(x+4,x+9), each with a difference of 5.

Thus sum of the last 5 terms = (x+5)+ (x+6)+ (x+7)+ (x+8)+ (x+9) = x+(5) + (x+1)+(5) + (x+2)+(5)+(x+3)+(5)+(x+4)+(5)

= x+(x+1)+(x+2)+(x+3)+(x+4)+5*5= 560+25 (as you are given that the sum of the 1st 5 terms of the sequence (=x+(x+1)+(x+2)+(x+3)+(x+4)) = 560.

Thus the sum of the last 5 terms = 585.

A is thus the correct answer.

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

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02 Feb 2016, 08:45
Dear Engr2012,

Sincere thanks

Engr2012 wrote:
studentsensual wrote:
thangvietnam
deepakpandey028

thx so much for ur incredible solution, cud i ask u to give a broader explanation?

thangvietnam wrote:
each of 5 last number is greater than each of 5 first number by 5.
we have 5 times 5=25

deepakpandey028 wrote:
560 + 5*5 = 585, Option A)

Assume that the terms of the sequence are

x
x+1
x+2
x+3
x+4

x+5
x+6
x+7
x+8
x+9

When you compare x with x+5 or x+1 with x+6, the difference is always = x+5-x = 5

Thus you can make 5 pairs
(x,x+5)
(x+1,x+6)
(x+2,x+7)
(x+3,x+8)
(x+4,x+9), each with a difference of 5.

Thus sum of the last 5 terms = (x+5)+ (x+6)+ (x+7)+ (x+8)+ (x+9) = x+(5) + (x+1)+(5) + (x+2)+(5)+(x+3)+(5)+(x+4)+(5)

= x+(x+1)+(x+2)+(x+3)+(x+4)+5*5= 560+25 (as you are given that the sum of the 1st 5 terms of the sequence (=x+(x+1)+(x+2)+(x+3)+(x+4)) = 560.

Thus the sum of the last 5 terms = 585.

A is thus the correct answer.

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

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07 Jun 2016, 10:01
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

Solution:

In solving this problem we must first remember that when we have 10 consecutive integers we can display them in terms of just 1 variable. Thus, we have the following:

Integer 1: x
Integer 2: x + 1
Integer 3: x + 2
Integer 4: x + 3
Integer 5: x + 4
Integer 6: x + 5
Integer 7: x + 6
Integer 8: x + 7
Integer 9: x + 8
Integer 10: x + 9

We are given that the sum of the first 5 integers is 560. This means that:

x + x+1 + x+2 + x+3 + x+4 = 560

5x + 10 = 560

5x = 550

x = 110

The sum of the last 5 integers can be expressed and simplified as:

x+5 + x+6 + x+7 + x+8 + x+9 = 5x + 35

Substituting 110 for x yields:

(5)(110) + 35 = 585

Alternatively, because both equations have 5x in common, we know that the difference between the sum of the first five numbers and the sum of the last five numbers is the difference between (1+2+3+4) and (5+6+7+8+9). Since 35 – 10 = 25, the sum of the last 5 is 585, which is 25 more than 560.
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink]

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07 Jun 2016, 20:20
x=first integer in sequence
5x+10=sum of first five integers in sequence
5x+35=sum of last five integers in sequence
(5x+35)-(5x+10)=25
560+25=585
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Re: In an increasing sequence of 10 consecutive integers, the [#permalink]

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07 Jun 2016, 21:04
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

Sum of 1st 5 integers = 560.

Avg of 1st 5 integers = 112.

As these are consecutive integers, this is nothing but the third term .
This means 1st term: 110.

First Method: (by AP formula)
Sum of next 10 terms can be calculated by AP formula: (n/2)* (2a + (n-1) d); where d = common difference = 1; n = total terms = 10; a = first term = 110
5 * (2*110+ 9) = 1145
Sum of last 5 terms = 1145 - 560 = 585.

Second method (Brute force)
3rd term is 112, 5th term would be 114.

sum of 115+116+...+120 = This simply adds up to 585
Sum of next 10 terms can be calculated by AP formula: (n/2)* (2a + (n-1) d); where d = common difference = 1; n = total terms = 10; a = first term = 110
5 * (2*110+ 9) = 1145
Sum of last 5 terms = 1145 - 560 = 585.

Re: In an increasing sequence of 10 consecutive integers, the   [#permalink] 07 Jun 2016, 21:04
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