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# A sequence consists of N consecutive even integers in descending order

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Math Expert
Joined: 02 Sep 2009
Posts: 65062
A sequence consists of N consecutive even integers in descending order  [#permalink]

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01 Jun 2020, 08:45
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Difficulty:

65% (hard)

Question Stats:

62% (02:43) correct 38% (03:13) wrong based on 34 sessions

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A sequence consists of N consecutive even integers in descending order. If the first integer is N and the average of the first five is 1594 more than the last integer, then what is the value of N?

(A) 600
(B) 750
(C) 700
(D) 800
(E) 840

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Joined: 18 Jan 2020
Posts: 1080
Location: India
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Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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01 Jun 2020, 09:53
Since set consist of even consecutive no.
Set will be {N,N-2,N-4,..............,(-N+2)}

Let's solve by options
1) N = 600.
Set will be (600,598,596,594,..........upto 600 term).
The average of first 5 no. Is (600+598+596+594+592)/5=596
And last no. Is -598
[ An = A+(n-1)D
A600=600+(600-1)-2
A600=600+(599)*-2
A600=600-1198
A600=-598 ]
The difference between the average and last no is 596-(-598) => 1194≠1594 (option A rejected)

2) N = 750.
Set will be (750,748,746,744,..........upto 750 term).
The average of first 5 no. Is (750+748+746+744+742)/5=746
And last no. Is -748
[ An = A+(n-1)D
A750=750+(750-1)-2
A750=750+(749)*-2
A750=750-1498
A750=-748 ]
The difference between the average and last no is 746-(-748) => 1494≠1594 (option B rejected)

3) N = 700.
Set will be (700,698,696,694,..........upto 700 term).
The average of first 5 no. Is (700+698+696+694+692)/5=696
And last no. Is -698
[ An = A+(n-1)D
A700=700+(700-1)-2
A700=700+(699)*-2
A700=700-1398
A700=-698 ]
The difference between the average and last no is 696-(-698) => 1394=1594 (option C rejected)

4) N = 800.
Set will be (800,798,796,794,..........upto 800 term).
The average of first 5 no. Is (800+798+796+794+792)/5=796
And last no. Is -798
[ An = A+(n-1)D
A800=800+(800-1)-2
A800=800+(799)*-2
A800=800-1598
A800=-798 ]
The difference between the average and last no is 796-(-798) => 1594=1594 (option D is qualifying)

5) N = 840.
Set will be (840,838,836,834,..........upto 840 term).
The average of first 5 no. Is (840+838+836+834+832)/5=836
And last no. Is -838
[ An = A+(n-1)D
A840=840+(840-1)-2
A840=840+(839)*-2
A840=840-1678
A840=-838 ]
The difference between the average and last no is 836-(-838) => 1674≠1594 (option E rejected)

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Posts: 114
Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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01 Jun 2020, 11:49
1
First 5 numbers in sequence: N, N-2, N-4, N-6, N-8. Average will be N-4
Let Y be the last integer.

Given, N-4=Y+1594 => N = Y+1598 ---(1)
Again, Y=N+(N-1)(-2) => Y = 2-N ---(2)

Substitute (2) in (1), 2N = 1600 => N=800
GMATWhiz Representative
Joined: 07 May 2019
Posts: 802
Location: India
Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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01 Jun 2020, 12:33
Bunuel wrote:
A sequence consists of N consecutive even integers in descending order. If the first integer is N and the average of the first five is 1594 more than the last integer, then what is the value of N?

(A) 600
(B) 750
(C) 700
(D) 800
(E) 840

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Solution

• The sequence consists of N “consecutive even numbers” in descending order.
o It means every term, after the first term, is 2 less than the previous term.
o Hence, the given sequence is in A.P.
• Thus, the given sequence is: $$N, N-2, N-4, …………N + (N-1)*(-2)$$
• The average of first five terms of the given A.P. = third term of the given A.P. $$= N-4$$
• Therefore, $$N – 4 = N +(N-1)*(-2) + 1594$$
$$⟹ N- 4 = -N + 1596$$
$$⟹2N = 1600$$
$$⟹ N = 800$$
Thus, the correct answer is Option D.
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Manager
Joined: 24 Sep 2014
Posts: 73
Concentration: General Management, Technology
Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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02 Jun 2020, 09:32
1
A sequence consists of N consecutive even integers in descending order. If the first integer is N and the average of the first five is 1594 more than the last integer, then what is the value of N?

Given:
$$N-2*(1-1), N-2*(2-1), N-2*(3-1), N-2*(4-1), N-2*(5-1),............ N-2*(N-1)$$, N consecutive even integers in descending order, with N as the first integer
Now, average of first five even integers is:
$$\frac{N-2*(1-1)+N-2*(2-1)+N-2*(3-1)+N-2*(4-1)+N-2*(5-1)}{5} = N-4$$ which is equal to 1594+last integer (N-2*(N-1))
$$N-4 = 1594+(-N+2) => N = 800$$
IESE School Moderator
Joined: 11 Feb 2019
Posts: 308
Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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03 Jun 2020, 13:54
IMO D

Tota numbers = N
First number = N
Since even consecutive numbers in decreasing order, common difference d = -2
First-term = N
Second Term = N-2
Third Term = N-4
Fourth Term = N-6
Fifth Term = N-8

Last term = N + (N-1)*(-2) (Using formula last term. = first term + (count -1)* comon differnece)

Given: average of the first five is 1594 more than the last integer

i.e.{N + (N-2) + (N-4) + (N-6) + (N-8)} /5 = N + (N-1)*(-2) + 1594

=>(5N-20)/5 = 1594 -N +2
= > N - 4 = 1596 -N
=> N = 800
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NJ
Re: A sequence consists of N consecutive even integers in descending order   [#permalink] 03 Jun 2020, 13:54