In an increasing sequence of 10 consecutive integers, the

Question

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

Bunuel · Accepted Answer

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.

thangvietname · Answer

each of 5 last number is greater than each of 5 first number by 5.

we have 5 times 5=25

agrchandan · Answer

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.

PareshGmat · Answer

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

Enael · Answer

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.

millopezle · Answer

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

studentsensual · Answer

thangvietnam
deepakpandey028

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

ENGRTOMBA2018 · Answer

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.

studentsensual · Answer

Dear Engr2012, Sincere thanks You've been very helpful

ScottTargetTestPrep · Answer

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.

gracie · Answer

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

mean of first five integers=560/5=112
because each of last five integers increases by 5,
mean of last five integers=117
and sum of last five integers=117*5=585
A

adiagr · Answer

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.

Alecita · Answer

AVG = Sum / # Terms

The sum of the 1st 5 integers = 560

The average of the 1st 5 integers = 560/5 = 112

The sequence of the 1st 5 integers is 110, 111, 112, 113, 114.

The sequence of the last 5 integers is 115, 116, 117, 118, 119

The sum of the last 5 integers is 117 x 5 = 585

Hence, answer A

* Note that the AVG = MEDIAN for Consecutives Only!!

I hope this help!

Thanks, Alecita

SchruteDwight · Answer

This is not true.

SchruteDwight · Answer

Multiple approaches to this question are possible. I think the easiest way to find the smallest term, however is as follows:

x=smallest term in the sequence –> 5x+(4(5))/2=560 (n(n+1)/2 is the formula for the sum of consecutive integers. 5 consecutive integers means x+x+1+x+2+x+3+x+4, hence it is a sum of (n-1) terms.

GmatKnightTutor · Answer

GMAT Problem Solving - Why adding with an algebra mindset can be helpful

https://www.youtube.com/watch?v=6rliv0d1uys

BrentGMATPrepNow · Answer

Here's another approach: 
Let a, b, c, d, and e be the first 5 consecutive integers, which means   a + b + c + d + e = 560

Here's a unique way to list all 10 consecutive integers:  a, b, c, d, e, a+5, b+5, c+5, d+5, e+5

So, the sum of the last 5 integers  = (a +5 ) + (b +5 ) + (c +5 ) + (d +5 ) + (e +5 ) 
 = a + b + c + d + e +  25  
 =  560  +  25  
 = 585

Answer: A

Abhishek009 · Answer

a + (a + 1) + (a + 2)+ (a + 3) + (a + 4) = 560

Thus,  5a = 550

Hence, a = 110

Thus, sum of last 5 numbers are :  115 + 116 + 117 + 118 + 119 = 585 , Answer will be (A)

Chibvike · Answer

Let the sequence of 10 consecutive integers be

x,x+1,x+2,...,x+9x.

The sum of the first 5 integers is given by: 
 
x+(x+1)+(x+2)+(x+3)+(x+4)=5x+10 
 
Given that this sum is 560, we can set up the equation: 
 5x+10=560 
 Solving for x: 
 5x=550 ⟹ x=1105x=550⟹x=110

The sum of the last 5 integers is: 
 (x+5)+(x+6)+(x+7)+(x+8)+(x+9)=5x+35 
 
Substituting x=110x=110: 
 5(110)+35=550+35=585 
 The correct answer is: 
  (A) 585

In an increasing sequence of 10 consecutive integers, the

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