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A sequence consists of N consecutive even integers in descending order

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

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New post 01 Jun 2020, 08:45
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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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Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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New post 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)

Answer is D

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

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

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

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New post 02 Jun 2020, 09:32
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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?

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))
\(N-4 = 1594+(-N+2) => N = 800 \)
Answer D
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Re: A sequence consists of N consecutive even integers in descending order  [#permalink]

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New post 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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Re: A sequence consists of N consecutive even integers in descending order   [#permalink] 03 Jun 2020, 13:54

A sequence consists of N consecutive even integers in descending order

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