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

BTGmoderatorLU wrote:Source: Economist GMAT

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

Say the first term = a; thus, the 2nd term = a – 7; the 3rd term = a – 7 – 7 = a – 2*7; the 4th term = a – 3*7;

Thus, the 40th term = a – 39*7

Since terms are decreasing in value, the smallest term would be the 40th term and the greatest term would be the first term = a = 281.

Thus, the smallest term = a – 39*7 = 281 – 273 = 8

The correct answer: A

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GRE Manhattan | ACT Tutoring San Antonio | GRE Prep Courses Boston | Houston IELTS Tutoring | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

BTGmoderatorLU wrote:Source: Economist GMAT

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

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