A sequence consists of N consecutive even integers in descending order

Question

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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Lipun · Accepted Answer

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

yashikaaggarwal · Answer

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 
 
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 
 
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 
 
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 
 
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 
 
The difference between the average and last no is 836-(-838) => 1674≠1594 (option E rejected)

Answer is D

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GMATWhizTeam · Answer

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.

kris19 · Answer

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

NitishJain · Answer

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

Kinshook · Answer

Given: A sequence consists of N consecutive even integers in descending order. 
Asked: 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?

The sequence = {N, N-2, N-4,....., N-(N-1)2=N-2N+2 = 2-N}
Average of first 5 ={ N+(N-2)+(N-4)+(N-6)+(N-8)}/5 = (N-4)
N-4 - (2-N) = 1594 
2N -6 = 1594
N = 800

IMO D

TithiMarodia · Answer

Hey, I did it in a different way, please correct me if I'm wrong somewhere

The sequence is N, N-2,N-4,N-6, N-8........N-2(x-1)
(x denotes the position of the number, for example - the first number = N-2(1-1) = N, second number = N-2(2-1)= n-2...) 
Therefore-
the average of first five numbers is N-4
N-4 = 1594 + N-2(x-1)
2x = 1600
x = 800

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A sequence consists of N consecutive even integers in descending order

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