Events & Promotions
|
|
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.
Jun 2106:30 AM PDT
-08:30 AM PDT
Not only the typical inference questions but also all other CR questions on the GMAT require the ability to deduce from the given information. Come master how to draw inferences accurately and efficiently with Verbal Expert-Sunita Singhvi. Limited Seats!
Jun 2111:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
Jun 2211:30 AM EDT
-12:30 PM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Jun 2308:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jun 3008:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates!
02 Apr 2020, 03:05
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
52% (02:25) correct
48%
(02:30)
wrong
based on 172
sessions
History
Date
Time
Result
Not Attempted Yet
Consider a function f satisfying \(f(x + y) = f(x)f(y)\) where x, y are positive integers, and \(f(1) = 2\). If \(f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)\) then \(a\) is equal to
A. 1
B. 2
C. 3
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
A. 1
B. 2
C. 3
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
Updated on: 18 Jun 2024, 08:16
Originally posted by GMATinsight on 02 Apr 2020, 03:37.
Last edited by Bunuel on 18 Jun 2024, 08:16, edited 4 times in total.
Last edited by Bunuel on 18 Jun 2024, 08:16, edited 4 times in total.
Kudos
Bookmarks
Bunuel
Consider a function f satisfying \(f(x + y) = f(x)f(y)\) where x, y are positive integers, and \(f(1) = 2\). If \(f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)\) then \(a\) is equal to
A. 1
B. 2
C. 3
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
\(f(x + y) = f(x)f(y)\)A. 1
B. 2
C. 3
D. 4
E. 5
Are You Up For the Challenge: 700 Level Questions
i.e. \(f(a + 1) = f(a)f(1) = 2f(a)\)
i.e. \(f(a + 2) = f[(a +1)+ 1] = f(a+1)f(1) = 4f(a) = 2^2f(a)\)
i.e. \(f(a + 3) = f[(a +2)+ 1] = f(a+2)f(1) = 8f(a) = 2^3f(a)\)
and so on
\(f(a + 1) + f(a + 2) + … + f(a + n) = 2f(a)*[1+2+2^2+2^3+---+2^n] = 16(2^n − 1)\)
Sum of n terms of a Geometric Progression
\(a+ar+ar^2+ar^3+---+ar^n = \frac{a(r^n-1)}{(r-1)}\)
\(a+ar+ar^2+ar^3+---+ar^n = \frac{a(r^n-1)}{(r-1)}\)
i.e. \(2f(a)*[1+2+2^2+2^3+---+2^n] = 2f(a)*[1*(2^n-1)/(2-1) = 16(2^n − 1)\)
i.e. \(2f(a) = 16\)
i.e. \(f(a) = 8\)
Also, \(f(2) = f(1)*(f1) = 2*2 = 4\)
\(f(3) = f(2)*(f1) = 4*2 = 8\)
i.e. \(f(3) = f(a)\)
i.e. a = 3
Answer: Option C
02 Apr 2020, 03:39
Kudos
Bookmarks
Consider a function f satisfying \(f(x + y) = f(x)f(y)\) where x, y are positive integers, and \(f(1) = 2\). If \(f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)\) then \(a\) is equal to
Let’s get some values of f(2), f(3),....
\(f(1) = 2^{1}\)
\(f(2) = f( 1+1) = 2*2= 2^{2}\)
\(f(3) = f( 1+2) = 2*2^{2} = 2^{3}\)
.....
\(f(n) = ....= 2^{n}\)
Well, \(f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)\)
—> \(f(a)f(1) + f(a)f(2) +...+f(a)f(n)\) =
\(f(a) 2^1 + f(a) 2^{2} +...+f(a) 2^{n}\)
—>\( f(a) ( 2^{1} + 2^{2} +...+2^{n} )\)=
\(f(a) ( 2( 2^n —1) / (2–1) ) = 16 (2^n—1)\)
—>\( f(a) = 8\)
Only f(3) satisfies that equation
—>\( a= 3\)
Answer (C)
Posted from my mobile device
Let’s get some values of f(2), f(3),....
\(f(1) = 2^{1}\)
\(f(2) = f( 1+1) = 2*2= 2^{2}\)
\(f(3) = f( 1+2) = 2*2^{2} = 2^{3}\)
.....
\(f(n) = ....= 2^{n}\)
Well, \(f(a + 1) + f(a + 2) + … + f(a + n) = 16(2^n − 1)\)
—> \(f(a)f(1) + f(a)f(2) +...+f(a)f(n)\) =
\(f(a) 2^1 + f(a) 2^{2} +...+f(a) 2^{n}\)
—>\( f(a) ( 2^{1} + 2^{2} +...+2^{n} )\)=
\(f(a) ( 2( 2^n —1) / (2–1) ) = 16 (2^n—1)\)
—>\( f(a) = 8\)
Only f(3) satisfies that equation
—>\( a= 3\)
Answer (C)
Posted from my mobile device
