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In an increasing set of 7 consecutive integers, the sum of t [#permalink]

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05 Nov 2012, 11:02

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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.

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.

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