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