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In an increasing sequence of 8 consecutive even integers, the sum of
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05 Apr 2017, 04:53
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In an increasing sequence of 8 consecutive even integers, the sum of the first 4 terms of the sequence is 124. What is the sum of the last 4 terms of the sequence? A. 136 B. 140 C. 144 D. 156 E. 160
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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05 Apr 2017, 05:55
Let Consecutive even integers be = 2n, 2n+2, 2n+4, 2n+6, ..... 2n+14 Sum of first four terms = 124. 8n+12 = 124 8n= 112. Sum of last 4 terms = 8n+44 = 112+44 = 156.
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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05 Apr 2017, 07:53
n + n + 2 + n + 4 + n + 6 = 124 4n + 12 = 124 4n/4 = 112/4 n = 28
28 + 8 + 28 + 10 + 28 + 12 + 28 + 14 156
Answer is D



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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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11 Apr 2017, 12:44
Bunuel wrote: In an increasing sequence of 8 consecutive even integers, the sum of the first 4 terms of the sequence is 124. What is the sum of the last 4 terms of the sequence?
A. 136 B. 140 C. 144 D. 156 E. 160 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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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07 Aug 2017, 03:01
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.



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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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07 Aug 2017, 13:14
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.
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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10 Aug 2017, 10:40
Bunuel wrote: In an increasing sequence of 8 consecutive even integers, the sum of the first 4 terms of the sequence is 124. What is the sum of the last 4 terms of the sequence?
A. 136 B. 140 C. 144 D. 156 E. 160 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
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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10 Aug 2017, 14:30
Bunuel wrote: In an increasing sequence of 8 consecutive even integers, the sum of the first 4 terms of the sequence is 124. What is the sum of the last 4 terms of the sequence?
A. 136 B. 140 C. 144 D. 156 E. 160 4n+12=124 n=28 n+7=35=mean 8*35124=156 D



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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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10 Aug 2017, 15:42
formula of sum for n of arithmetic progression sum = n/2(2x + (n1)*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



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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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11 Aug 2017, 06:16
Hi Bunuel, If we consider the even consecutive intergers as 2a8, 2a6, 2a4, 2a2, 2a+2, 2a+4, 2a+6, 2a+8, then 8a20 = 124 ==> a= 18 and 8a+20 ==> 8x18+20= 164. I think the question is for consecutive positive even integers.



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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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11 Aug 2017, 06:28
shivam2506 wrote: Hi Bunuel, If we consider the even consecutive intergers as 2a8, 2a6, 2a4, 2a2, 2a+2, 2a+4, 2a+6, 2a+8, then 8a20 = 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: 2a8, 2a6, 2a4, 2a2, 2a, 2a+2, 2a+4, 2a+6. So, we need 8a + 12, which is 156.
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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11 Aug 2017, 08:19
n(2a+(n1)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 + (81)2 )/2 = 270 S1= 124 given S = S2 S1 = 270124 = 156. Option D
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Re: In an increasing sequence of 8 consecutive even integers, the sum of
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