The largest number in a series of consecutive even integers

Question

The largest number in a series of consecutive even integers is w. If the number of integers is n, what is the smallest number in terms of w and n?

A. w– 2n
B. w–n + 1
C. w– 2(n– 1)
D. n– 6 + w
E. w – n/2

(A): w– 2n = 10 – 2(2) = 10 – 4 = 6 ≠ 8
(B): w–n + 1 =10 – 2 +1 = 9 ≠ 8
(C): w– 2(n– 1) = 10 – 2(2 – 1) = 10 –2 = 8 = 8
(D): n– 6 + w = 2 – 6 + 10 = 6 ≠ 8
(E): w – n/2 = 10 – 2/2 = 10 – 1 = 9 ≠ 8

Only (C) matches target number.
Conceptual approach is a bit trickier:
In a set of n consecutive integers in descending order, series would be {w, w-1, w-2, w-3...} - that is, each subsequent integer is w minus integer's position in sequence below w, or w - (n - 1).
For example, set {4, 3, 2, 1} may be written as {4, (4 -1), (4 - 2), (4 - 3)}.

Since this is a series of EVEN numbers, next smaller number in series is w - 2, and next is w - 4 - in other words, each subsequent integer subtracts 2 times number of its position below w, or w - 2(n -1).
For example, set {8, 6, 4} may be written as {8, (8 - 2(2-1), (8 - 2(3-1)}.

KarishmaB · Accepted Answer

Another method would be to use the concept of Arithmetic Progressions. Consecutive even integers form an AP where the common difference is 2.
In an AP, Last term = First term + (n-1)*common difference
w = First term + (n-1)*2
First term = w - 2(n-1)

tatachalikyan · Answer

Ciao

I would pick a random set of 3 consecutive even integers e.g. {2, 4, 6} and try all the answers choices. 
IMO will be quicker.

Thanx
TC

goodyear2013 · Answer

{2, 4, 6} --> i.e. W = 6, n = 3 to get smallest one; 2
6 - 6 = 0
6 - 3 + 1 = 4
6 - 2(2) = 2 --> Answer
3 - 6 + 6 = 3
6 - 3/2 = 4.5

Cool. Thanks!

CrackverbalGMAT · Answer

The largest number in a series of consecutive even integers is w. If the number of integers is n, what is the smallest number in terms of w and n?

w– 2n
w–n + 1
w– 2(n– 1)
n– 6 + w
w – n/2

Let us say the consecutive even integers are 0, 2, 4 where w = 4, n = 3

By plugging in these values our answer should be 0

(A) 4 - 6
(B) 4 - 3 + 1
(C) 4 - 2(2)
(D) 3 - 6 + 4
(E) 4 - 3/2

Only option C satisfies that

reachasi · Answer

Since the integers are even numbers, let's assume that the first term is 2k. As the series is consecutive even integers, its 2nd term will be 2k+2, 3rd term will be 2K+4, and so on thus giving the nth term as 2k + 2(n-1).

Now,

2k+2(n-1) = w

or, 2k = w-2(n-1)

Now,

2k+2(n-1) = w

or, 2k = w-2(n-1)

thesip2323 · Answer

Don't know if my method is the fastest way to solve it but I'll note it down anyway (TTP way).

Let x = smallest value and thus the value we are solving for

(W-X)/2 + 1 = N
W-X/2 = N-1
W-X = 2N - 2
-X = 2N-2-W
X = -(2N-2-W)
X = -2N+2+W

Now rearrange in order of answer choices (seems like W is the first element in choices), so:
X = W-2N+2 
X=W-2(N-1)

bumpbot · Answer

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The largest number in a series of consecutive even integers

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The largest number in a series of consecutive even integers

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