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In a sequence of 40 numbers, each term, except for the first

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by AAPL » Tue Jun 26, 2018 5:55 am

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Answers

A

B

C

D

E

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Difficulty—

In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. -6

The OA is A.

I think it should be A.

The greatest term is 281 which would also be the first term.

Now, we have to find the last term that is least among all.

Difference between first and last = 39*7 = 273.

So, the last term is 281 - 273 = 8.

Has anyone another strategic approach to solve this PS question? Regards!
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by Brent@GMATPrepNow » Tue Jun 26, 2018 12:34 pm
AAPL wrote:In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. -6
Notice that each term is 7 less than the term before it.
This means the greatest term will be the first term

In other words, term1 = 281

Let's list a few terms of the sequence and look for a pattern:
term1 = 281
term2 = 281 - 7
term3 = 281 - 7 - 7 = 218 - (2)(7)
term4 = 281 - 7 - 7 - 7 = 218 - (3)(7)
term5 = 281 - 7 - 7 - 7 - 7 = 218 - (4)(7)
term6 = 281 - 7 - 7 - 7 - 7 - 7 = 218 - (5)(7)
.
.
.
term40 = 281 - 7 - 7 - 7 - 7 - 7 - 7 . . . = 281 - (39)(7)

As we can see, term40 wil be the smallest term.
So, let's evaluate term40

term40 = 281 - (39)(7)
= 281 - 273
= 8

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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