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In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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05 Nov 2012, 12:02
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In an increasing set of 7 consecutive integers, the sum of the last 4 integers is 34. What is the sum of the first 3 integers in the set? A. 12 B. 15 C. 18 D. 36 E. 120 Now my doubt is: A question from OG 13 page num 164 ( Q no. 90), this question i solved in the following manner Sum of (1 to 5) = 15 Sum of last 5 (6 to 10) is 40 Differnce is 25 As per question we know sum of first 5 so we can obtain sum of last 5 by adding 25 to it. The method is also explained as alternative method in the OG If the same concept we apply in the problem mentioned above Sum of (1,2,3) = 6 Sum of (4,5,6,7) = 22 difference is 16 So if we subtract 16 from 34 ie 18 should be the answer. One difference her i found is that the difference of 16 is not common. So what is the principle and logic here. Method x,x+1.... is known to me Pls explain my doubt.
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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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05 Nov 2012, 12:11
Archit143 wrote: In an increasing set of 7 consecutive integers, the sum of the last 4 integers is 34. What is the sum of the first 3 integers in the set? A. 12 B. 15 C. 18 D. 36 E. 120 Now my doubt is: A question from OG 13 page num 164 ( Q no. 90), this question i solved in the following manner Sum of (1 to 5) = 15 Sum of last 5 (6 to 10) is 40 Differnce is 25 As per question we know sum of first 5 so we can obtain sum of last 5 by adding 25 to it. The method is also explained as alternative method in the OG If the same concept we apply in the problem mentioned above Sum of (1,2,3) = 6 Sum of (4,5,6,7) = 22 difference is 16 So if we subtract 16 from 34 ie 18 should be the answer. One difference her i found is that the difference of 16 is not common. So what is the principle and logic here. Method x,x+1.... is known to me Pls explain my doubt. I'm happy to help with this. The trouble is: this trick (finding the sum of the last set minus the sum of the first set) only works if the two sets are of equal size. Once the sets are of unequal sizes, the difference will change radically  think if the first three integers are {101, 102, 103} and the last four are {104, 105, 106, 107}  the the sum of the latter will be over a hundred more than the sum of the former! I would say  it's a good start to add 4+5+6+7 = 22. We know that 34 is 12 more than 22, so that means each of the four terms needs to be 12/4 = 3 units bigger  instead of {4, 5, 6, 7}, those last four will be "up three" from those: {7, 8, 9, 10}. Then the first three are {4, 5, 6}, and their sum if 15. Does this make sense? Mike
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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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05 Nov 2012, 13:45
x + (x+1) + ... => 7x + 21 given: 4x + 18 = 34 x = 4
4(3) + 3 = 15



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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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05 Nov 2012, 13:46
any questions, please ask



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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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05 Nov 2012, 13:49
Thanks Mike
excellent way of solving, explanation in the last part is really good.



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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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05 Nov 2012, 13:51
watwazdaquestion wrote: any questions, please ask pls go through my doubt mentioned in my question



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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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09 Dec 2012, 20:46
7 consecutive integers is similar to : {x,x+1,x+2} {x+3,x+4,x+5,x+6} Sum of last 4: 4x + 18 = 34 ==> 4x = 16 ==> x= 4 Sum of first 3: 3x + 3 = 3 (4) + 3 = 15 Answer: B
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Re: In an increasing set of 7 consecutive integers, the sum of t [#permalink]
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08 Mar 2016, 07:58
Archit143 wrote: watwazdaquestion wrote: any questions, please ask pls go through my doubt mentioned in my question Here let the first integer be x => 4x+28 =34 => x = 4 hence => sum = 4+5+6 => 15
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Re: In an increasing set of 7 consecutive integers, the sum of t
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