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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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In an increasing sequence of 10 consecutive integers, the
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07 Dec 2012, 03:38
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
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06 Feb 2013, 02:05
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
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06 Feb 2013, 02:23
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.



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Re: In an increasing sequence of 10 consecutive integers, the
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25 Mar 2014, 00:26
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
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20 Jul 2014, 19:26
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
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19 Feb 2015, 14:27
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(101))=5(229)=1145 Total sum sum of first five terms=1145560=585



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Re: In an increasing sequence of 10 consecutive integers, the
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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
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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+5x = 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
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02 Feb 2016, 08:45
Dear Engr2012, Sincere thanks You've been very helpful 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+5x = 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
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07 Jun 2016, 10:01
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.
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In an increasing sequence of 10 consecutive integers, the
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Updated on: 02 Oct 2018, 16:10
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
Originally posted by gracie on 07 Jun 2016, 20:20.
Last edited by gracie on 02 Oct 2018, 16:10, edited 1 time in total.



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Re: In an increasing sequence of 10 consecutive integers, the
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07 Jun 2016, 21:04
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 + (n1) 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 + (n1) 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.



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Re: In an increasing sequence of 10 consecutive integers, the
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02 Oct 2018, 15:18
AVG = Sum / # Terms The sum of the 1st 5 integers = 560The 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 = 585Hence, answer A * Note that the AVG = MEDIAN for Consecutives Only!!I hope this help! Thanks, Alecita




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