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sum of sequence

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by Patrick_GMATFix » Wed Jan 29, 2014 8:11 am
Since we're dealing with consecutive integers, the 6th is 5 more than the 1st, the 7th is 5 more than the 2nd etc... Since each unknown integer is 5 more than each known integer, the five unknown integers will have a sum 5*5=25 more than the sum of the five known integers. The sum of the last five integers must be 560 + 25 = 585. The full solution below is taken from the GMATFix App.

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by Brent@GMATPrepNow » Wed Jan 29, 2014 8:21 am
kobel51 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
Patrick's solution is definitely the fastest.
That said, here's another (slightly longer, but still effective :-)) approach:

Let x = 1st number
So, x+1 = 2nd number
x+2 = 3rd number
x+3 = 4th number
.
.
.
x+8 = 9th number
x+9 = 10th number

The sum of the first 5 integers is 560
x + (x+1) + (x+2) + (x+3) + (x+4) = 560
5x + 10 = 560
5x = 550
x = 110 = 1st term (111 = 2nd term, 112 = 3rd term, etc)

So, 6th term = 115, 7th term = 116, . . . 10th term = 119,

What is the sum of the last 5 integers in the sequence?
115 + 116 + 117 + 118 + 119 = [spoiler]585 = A[/spoiler]

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by GMATGuruNY » Wed Jan 29, 2014 1:29 pm
kobel51 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
With evenly spaced integers:
Average = median = sum/number.
Sum = number*median.

In the problem at hand:
The median of the first five integers = sum/number = 560/5 = 112.
Thus, the 10 integers are as follows:
110, 111, 112, 113, 114
115, 116, 117...

The median of the last five integers is 117.
Thus:
Sum of the last five integers = number*median = 5*117 = 585.

The correct answer is A.
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by rairavig » Wed Jan 29, 2014 8:21 pm
difference between any 1st and 6th consecutive integer is 5
difference between any 2nd and 7th consecutive integer is again 5
so the total difference will be 5*5=25
Ans is 560+25=585
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by Jeff@TargetTestPrep » Tue Jun 23, 2015 2:16 pm
kobel51 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
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

Answer: A

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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Head of GMAT Instruction
[email protected]

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by Matt@VeritasPrep » Tue Jun 23, 2015 5:56 pm
Another option here: we've got a set of consecutive integers, so we know that the middle number = sum/(# of terms).

Given 5 integers with a sum of 560, the middle number is thus (560 / 5), or 112.

To get to the middle of the NEXT five integers, we just add 5. That means the middle number of the next five is 117, at which point we can work backwards: multiply this by 5, and we have a sum of 5*117, or 585.
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