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Archit143
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In an increasing set of 7 consecutive integers, the sum of 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

Show Spoiler
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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B
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mikemcgarry
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In an increasing set of 7 consecutive integers, the sum of 05 Nov 2012, 12:11
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Archit143
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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In an increasing set of 7 consecutive integers, the sum of 05 Nov 2012, 13:45
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x + (x+1) + ... => 7x + 21
given: 4x + 18 = 34
x = 4

4(3) + 3 = 15
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In an increasing set of 7 consecutive integers, the sum of 09 Dec 2012, 20:46
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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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In an increasing set of 7 consecutive integers, the sum of 23 Feb 2022, 04:46
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4 integers(in the increasing order) sum is 34 = 34/4 = 8.5 = 7+8+9+10 (as 8.5 is the mean of these 4 numbers)

Then the first 3 integers must be 4+5+6 (in the increasing order) = 15 ---> Correct answer is "B"
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In an increasing set of 7 consecutive integers, the sum of 23 Feb 2022, 06:24
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Let n, n+1,n+2....n+6 be the seven consecutive integers.

Sum of last 4 consecutive integers = 34

(n+3)+(n+4)+(n+5)+(n+6) = 34
SO n = 4.

Sum of first 3 consecutive integers ?

n + (n+1)+(n+2) = 3n+3 = 3(4) +3 = 15
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In an increasing set of 7 consecutive integers, the sum of 28 Apr 2022, 11:57
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let 7 consecutive numbers be a,b,c,d,e,f,g

sum of last 4 digits be d+e+f+g

since the digits are consecutive we can write them as d+d+1+d+2+d+3=34
solving this d=7
abc are 4,5,6 their sum is 15
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In an increasing set of 7 consecutive integers, the sum of 24 Jul 2024, 00:57
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Sum of 7 consecutive integers:
x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6)

Sum of last four integers:
(x+3) + (x+4) + (x+5) + (x+6) = 34
4x + 18 = 34
4x = 16
x = 4

Sum of first 3 integers:
x + (x+1) + (x+2)
4 + 5 + 6 = 15

Option B

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In an increasing set of 7 consecutive integers, the sum of 11 Jan 2025, 05:57
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Fourth term of the 7 consecutive integers is the mean. Let it be m. 7th term = m+3
Equating averages
(m+m+3)/2 = 34/4
2m+ 3 = 17
2m = 14
m = 7

Overall sum = 7*7 (terms * average)
Sum of First 3 = 49-34 = 15
Archit143
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

Show Spoiler
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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