|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
16 Sep 2014, 01:25
Kudos
Bookmarks
In the infinite sequence 1, 3, 9, 27, ... , each term after the first is three times the preceding term. What is the positive difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms of the sequence?
A. \(10*3^{11}\)
B. \(20*3^{11}\)
C. \(10*3^1\)
D. \(40*3^{11}\)
E. \(20*3^{12}\)
A. \(10*3^{11}\)
B. \(20*3^{11}\)
C. \(10*3^1\)
D. \(40*3^{11}\)
E. \(20*3^{12}\)
16 Sep 2014, 01:25
Kudos
Bookmarks
Official Solution:
In the infinite sequence 1, 3, 9, 27, ... , each term after the first is three times the preceding term. What is the positive difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms of the sequence?
A. \(10*3^{11}\)
B. \(20*3^{11}\)
C. \(10*3^1\)
D. \(40*3^{11}\)
E. \(20*3^{12}\)
You don't need to use the geometric progression formula to solve this question. Instead, simply recognize the pattern:
\(b_1=1=3^0\);
\(b_2=3=3^1\);
\(b_3=9=3^2\);
\(b_4=27=3^3\);
...
Thus, for any term \(b_n\), its value is \(3^{n-1}\).
The difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms is:
\((b_{13}+b_{15})-(b_{12}+b_{14})=\)
\(=3^{12}+3^{14}-3^{11}-3^{13}=\)
\(=3^{11}(3+3^3-1-3^2)=\)
\(=20*3^{11}\)
Answer: B
In the infinite sequence 1, 3, 9, 27, ... , each term after the first is three times the preceding term. What is the positive difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms of the sequence?
A. \(10*3^{11}\)
B. \(20*3^{11}\)
C. \(10*3^1\)
D. \(40*3^{11}\)
E. \(20*3^{12}\)
You don't need to use the geometric progression formula to solve this question. Instead, simply recognize the pattern:
\(b_1=1=3^0\);
\(b_2=3=3^1\);
\(b_3=9=3^2\);
\(b_4=27=3^3\);
...
Thus, for any term \(b_n\), its value is \(3^{n-1}\).
The difference between the sum of the 13th and 15th terms and the sum of the 12th and 14th terms is:
\((b_{13}+b_{15})-(b_{12}+b_{14})=\)
\(=3^{12}+3^{14}-3^{11}-3^{13}=\)
\(=3^{11}(3+3^3-1-3^2)=\)
\(=20*3^{11}\)
Answer: B
23 Oct 2023, 03:36
Kudos
Bookmarks
I've revised the question and solution, incorporating additional details for improved clarity. I trust this makes it more comprehensible.
