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
[spoiler]OA=A[/spoiler].
What is the best approach I can use here? Could anyone give me some help? Please. I'd be thankful.
In a sequence of 40 numbers, each term, except for
This topic has expert replies
-
- Legendary Member
- Posts: 1622
- Joined: Thu Mar 01, 2018 7:22 am
- Followed by:2 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
-
- Legendary Member
- Posts: 2898
- Joined: Thu Sep 07, 2017 2:49 pm
- Thanked: 6 times
- Followed by:5 members
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 (nº40, 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.
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 (nº40, 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.
- 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
Since each term is 7 less than the preceding term, the numbers are EVENLY SPACED.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. −6
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‚€ = ?
a� = 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
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
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7294
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
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.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. −6
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]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews