In an increasing sequence of 8 consecutive even integers, the sum of

We can represent the integers as follows:

x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12, x + 14

Since the sum of the first 4 terms is 124:

x + x + 2 + x + 4 + x + 6 = 124

4x + 12 = 124

4x = 112

x = 28

Finally, since the last 4 terms are x + 8, x + 10, x + 12 and x + 14, the sum of the these 4 numbers is:

(28 + 8) + (28 + 10) + (28 + 12) + (28 + 14) = 36 + 38 + 40 + 42 = 156

Alternative solution:

Recall that we can represent the the integers as:

x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12, x + 14

We can see that the fifth integer, x + 8, is 8 more than the first integer, x. Likewise, each of the sixth, seventh and eighth integers is 8 more than the second, third and fourth integers. Thus, the sum of the last 4 integers should be 4(8) = 32 more than the sum of the first 4 integers. Since we are given that the sum of the first 4 integers is 124, the sum of the last 4 integers is 124 + 32 = 156.

Answer: D

Get first term from problem statement
a+a+2+a+4+a+6=124
a=28
Calculate t5 & t8 of the series -> calculate the avg
S4 of last 4 terms=n *avg of t5 & t8 =4*39=156.
Option D.

My method is a little different than those listed above:

The first four terms total 124, so the average of the first four terms is:
124/4 = 31
Given that the numbers are even and consecutive, you know the four numbers must straddle 31; two greater than, two less than: 28, 30, 32, and 34 (which total 124).

So the next four, consecutive numbers are: 36, 38, 40, and 42, totaling 156.

Knowing the sum of the first 4 terms is 124, we can create the following equation in which x = the first term:

x + x + 2 + x + 4 + x + 6 = 124

4x = 112

x = 28

Thus, the sum of the last 4 terms is:

x + 8 + x + 10 + x + 12 + x + 14 = 4x + 44 = 4(28) + 44 = 156

Alternate solution:

We can let these 8 terms be x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12, x + 14. Notice that each of the last four terms (in bold) is 8 more than each of the first four terms, respectively. Thus the sum of the last four terms must be 8(4) = 32 more than the sum of the first four terms. We are given that the sum of the first four terms is 124; thus, the sum of the last four terms must be 124 + 32 = 156.

Answer: D

let n=term 1
4n+12=124→
n=28
n+(3*2)+1=35=median of sequence
8*35-124=156=sum of last 4 terms
D

formula of sum for n of arithmetic progression sum = n/2(2x + (n-1)*d)
d = 2 (consecutive even integers)
n = 4 (first 4 integers)
x = first number in the progression

124 = 4/2(2x + (4 - 1)*2)
124= 2(2x + 6)
4x = 124 - 12
x= 28

so first 4 even consecutive integers are: 28, 30, 32, 34 = sum 124
then next 4 are: 36 + 38 + 40 + 42 = 156
Answer D

Hi Bunuel,

If we consider the even consecutive intergers as 2a-8, 2a-6, 2a-4, 2a-2, 2a+2, 2a+4, 2a+6, 2a+8, then
8a-20 = 124 ==> a= 18

and 8a+20 ==> 8x18+20= 164.

I think the question is for consecutive positive even integers.

You are missing 2a as the fifths term: 2a-8, 2a-6, 2a-4, 2a-2, 2a, 2a+2, 2a+4, 2a+6.

So, we need 8a + 12, which is 156.

n(2a+(n-1)d)/(2) = 124
n=4, d=2 as even integers
put the values get a=28
now same formula as above put n=8 and a=28
S2= 8(2x28 + (8-1)2 )/2 = 270
S1= 124 given
S = S2- S1 = 270-124 = 156. Option D

Question: I'm trying to wrap my head around why you can use n OR 2n to start the sequence. This tripped me up on my practice exam...should have just went with either. To clarify.....

Sum of first four integers...
x+x+2+x+4+x+6=124
4x+12=124
x=28

Sum of last four

x+8+x+10+x+12+x+14=?
4x+44=?
4(28)+44=156

Now starting with 2x...as I'm typing this out I realize that I'm still expressing consecutive even integers when I start with 2x but is there more to it?

Sum of first four integers...
2x+2x+2+2x+4+2x+6=124
8x+12=124
x=14

Sum of last four
2x+8+2x+10+2x+12+2x+14=?
4(2x)+44=?
4(2(14))+44=156

In both cases you get ans. D 156.

Bunuel you got anything?

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