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In an increasing sequence of 5 consecutive even integers, the sum of

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In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 07 Mar 2017, 18:22
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In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer
is 132. What is the sum of the first and last integers?

A.84
B.86
C.88
D.90
E.92
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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 07 Mar 2017, 23:54
BPHASDEU wrote:
In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer
is 132. What is the sum of the first and last integers?

A.84
B.86
C.88
D.90
E.92


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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 08 Mar 2017, 02:01
1
2
let the number be a-2, a-1, a, a+1, a+2
according to the question, a -1 +a +a +1 = 132
or 3a = 132
or a = 44
therefore first integer +last integer = a-2 +a+2 = 2a = 2*44 = 88

Option C
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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 08 Mar 2017, 21:16
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Let the 5 consecutive even numbers be x, x+2, x+4, x+6, x+8

Given: x+2+x+4+x+6 = 132
=> 3x+12 = 132
=> x = 40

Since x = 40; x+8 = 48

x + (x+8) = 40+48 = 88 (C)
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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 10 Mar 2017, 12:39
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Hi All,

This question can be solved with the Average Formula...

We're told that the sequence is 5 CONSECUTIVE EVEN integers and the sum of the 2nd, 3rd and 4th is 132. Since the numbers are CONSECUTIVE, the "middle" one will be the AVERAGE of the 2nd, 3rd and 4th.

Average = Sum/Number of terms = 132/3 = 44

Thus, since the 2nd, 3rd and 4th terms are 42, 44 and 46 respectively. By extension, the sum of the 1st and 5th terms is: 40 + 48 = 88

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In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 10 Mar 2017, 17:45
BPHASDEU wrote:
In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer
is 132. What is the sum of the first and last integers?

A.84
B.86
C.88
D.90
E.92


132/3=44=median
44-4=40
44+4=48
40+48=88
C
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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 18 Mar 2017, 18:01
BPHASDEU wrote:
In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer
is 132. What is the sum of the first and last integers?

A.84
B.86
C.88
D.90
E.92


let the numbers be a-2d,a-d,a,a+d,a+2d

sum of 2nd,3rd and 4th will be a-d+a+a+d=3a = 132 --> a=44

since the numbers are consecutive even integers d should be 2, so the numbers are 40,42,44,46,48

Answer C (48)
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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 06 Jul 2018, 08:22
BPHASDEU wrote:
In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer is 132. What is the sum of the first and last integers?

A.84
B.86
C.88
D.90
E.92


Let the numbers be " a , a +1 , a + 2 , a + 3 & a +4 "

Now, \(( a + 1 ) + ( a + 2 ) + ( a + 3 ) = 132\)

Or, \(3a + 6 = 132\)

Or, \(3a = 126\)

Or, \(a = 42\)

So, Sum of the first and last Integer must be 42 + 46 = 90, the answer must be (D)
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Re: In an increasing sequence of 5 consecutive even integers, the sum of  [#permalink]

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New post 13 Jul 2018, 14:15
BPHASDEU wrote:
In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer
is 132. What is the sum of the first and last integers?

A.84
B.86
C.88
D.90
E.92


Letting x denote the smallest of these even consecutive integers, we can create the equation:

x + 2 + x + 4 + x + 6 = 132

3x + 12 = 132

3x = 120

x = 40

So the sum of the first and last integers is 40 + (40 + 8) = 40 + 48 = 88.

Answer: C
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Re: In an increasing sequence of 5 consecutive even integers, the sum of &nbs [#permalink] 13 Jul 2018, 14:15
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