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

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

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

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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.

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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 pmFollowed by: 5 members; Thanked: 6 times

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.

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

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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

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