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