In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68...

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AAPL: Moderator; Posts: 2749; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

by AAPL » Sun Jul 12, 2020 7:57 am

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Princeton Review

In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68. What is the product of the last three integers in the sequence?

A. 1,000
B. 960
C. 925
D. 30
E. 25

OA B

Source: Beat The GMAT — Problem Solving | Sun Jul 12, 2020 7:57 am

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Re: In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68...

by Scott@TargetTestPrep » Sun Jul 19, 2020 10:36 am

AAPL wrote: ↑
Sun Jul 12, 2020 7:57 am
Princeton Review

In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68. What is the product of the last three integers in the sequence?

A. 1,000
B. 960
C. 925
D. 30
E. 25

OA B

Solution:

The seven even integers in the sequence are as follows:

x, x - 2, x - 4, x - 6, x - 8, x - 10, x - 12

Since the sum of the first 4 integers is 68, we have:

x + x - 2 + x - 4 + x - 6 = 68

4x = 80

x = 20

Since our last 3 integers are x - 8, x - 10, and x - 12, they are 12, 10, and 8, and thus their product is 12 x 10 x 8 = 960.

Answer: B

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