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

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

Gmat_mission: Legendary Member; Posts: 1622; Joined: Thu Mar 01, 2018 7:22 am; Followed by:2 members

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

by Gmat_mission » Mon Jun 25, 2018 12:03 am

Timer

00:00

Your Answer

Global Stats

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. âˆ’6

[spoiler]OA=A[/spoiler].

What is the best approach I can use here? Could anyone give me some help? Please. I'd be thankful.

Quote

Source: Beat The GMAT — Problem Solving | Mon Jun 25, 2018 12:03 am

Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Mon Jun 25, 2018 1:23 am

Hi Gmat_misson.

Let's take a look at your question.

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?

We have that $$a_1=281,\ a_2=281-7,\ \ a_3=\left(281-7\right)-7,\ a_4=\left(\left(281-7\right)-7\right)-7$$ $$a_1=281,\ a_2=a_1+\left(-7\right),\ \ a_3=a_1+\left(-14\right),\ a_4=a_1+\left(-21\right),\ ....$$ $$a_1=281,\ a_2=a_1+\left(-7\right),\ \ a_3=a_1+2\cdot\left(-7\right),\ a_4=a_1+\left(3\right)\left(-7\right),\ ....$$ Therefore, we get the general form: $$a_n=a_1+\left(n-1\right)\left(-7\right).$$ So, the last term (nÂº40, the smallest) is equal to $$a_{40}=a_1+\left(40-1\right)\left(-7\right)=281+\left(39\right)\left(-7\right)=281-273=8.$$ So, the correct answer is the option A.

I hope this explanation is clear enough.

Regards.

Quote

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon Jun 25, 2018 2:11 am

Gmat_mission 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. âˆ’6

Since each term is 7 less than the preceding term, the numbers are EVENLY SPACED.
For any evenly spaced set, where d is the distance between successive terms:
$$a_{n} = a_{1} - (n-1)d$$
In the problem above:
aâ‚„â‚€ = ?
aâ‚� = 281
n = 40
Since each term is 7 less than the preceding term, d =7.
Plugging these values into the equation above, we get:
aâ‚„â‚€ = 281 - (40-1)(7) = 281 - 273 = 8.

The correct answer is A.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Quote

GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue Jun 26, 2018 5:18 pm

Gmat_mission 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. âˆ’6

Since each term, except for the first one, is 7 less than the previous term, it must mean the first term is the greatest term, i.e., a_1 = 281. The smallest term must be the last term, i.e., the 40th term.

The sequence depicted is also an arithmetic sequence and we know that the nth term of an arithmetic sequence has the formula a_n = a_1 + d(n - 1). The variable d is the common difference and here the common difference is -7 (since each term is 7 less than the previous term). Thus, we have

a_40 = 281 + (-7)(40 - 1)

a_40 = 281 - 7(39)

a_40 = 281 - 273

a_40 = 8

Answer: A

Scott Woodbury-Stewart
Founder and CEO
[email protected]