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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.

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

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07 Jun 2016, 11:01

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

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. _________________

Jeffrey Miller Scott Woodbury-Stewart Founder and CEO

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

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07 Jun 2016, 22:04

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

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.

A is the answer.

gmatclubot

Re: In an increasing sequence of 10 consecutive integers, the
[#permalink]
07 Jun 2016, 22:04

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