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.
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2212:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Updated on: 28 Feb 2019, 00:54
Originally posted by EgmatQuantExpert on 22 Feb 2019, 04:52.
Last edited by EgmatQuantExpert on 28 Feb 2019, 00:54, edited 1 time in total.
Last edited by EgmatQuantExpert on 28 Feb 2019, 00:54, edited 1 time in total.
Kudos
Bookmarks
B
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:
40% (02:58) correct
60%
(02:45)
wrong
based on 212
sessions
History
Date
Time
Result
Not Attempted Yet
e-GMAT Question of the Week #37
The \(n^{th}\) term of a sequence \(a_1, a_2, a_3, … a_n\) is given by \(a_n = \frac{1}{n(n+2)}\). What is the sum of the first 20 terms of the sequence?
A. \(\frac{300}{440}\)
B. \(\frac{325}{462}\)
C. \(\frac{303}{462}\)
D. \(\frac{375}{450}\)
E. \(\frac{650}{462}\)
The \(n^{th}\) term of a sequence \(a_1, a_2, a_3, … a_n\) is given by \(a_n = \frac{1}{n(n+2)}\). What is the sum of the first 20 terms of the sequence?
A. \(\frac{300}{440}\)
B. \(\frac{325}{462}\)
C. \(\frac{303}{462}\)
D. \(\frac{375}{450}\)
E. \(\frac{650}{462}\)
22 Feb 2019, 06:18
Kudos
Bookmarks
EgmatQuantExpert
\(a_n = \frac{1}{n(n+2)}\)
\(a_1 = \frac{1}{1(1+2)} = \frac{1}{1*3} = \frac{1}{2}[(\frac{1}{1}) - (\frac{1}{3})]\)
\(a_2 = \frac{1}{2(2+2)} = \frac{1}{2*4}= \frac{1}{2}[(\frac{1}{2}) - (\frac{1}{4})]\)
\(a_3 = \frac{1}{3(3+2)} = \frac{1}{3*5}= \frac{1}{2}[(\frac{1}{3}) - (\frac{1}{5})]\)
\(a_4 = \frac{1}{4(4+2)} = \frac{1}{4*6}= \frac{1}{2}[(\frac{1}{4}) - (\frac{1}{6})]\)
\(a_5 = \frac{1}{5(5+2)} = \frac{1}{5*7}= \frac{1}{2}[(\frac{1}{5}) - (\frac{1}{7})]\)
and so on....
\(a_{20} = \frac{1}{20(20+2)} = \frac{1}{20*22}= \frac{1}{2}[(\frac{1}{20}) - (\frac{1}{22})]\)
Adding them we can see that all terms will be cancelled out except
Sum of 20 terms \(= (\frac{1}{2})[ (\frac{1}{1}) - \frac{1}{3} + (\frac{1}{2}) - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \frac{1}{5} - \frac{1}{7} + ------- +\frac{1}{16} - \frac{1}{18} + \frac{1}{17} - \frac{1}{19} + \frac{1}{18} - \frac{1}{20} + \frac{1}{19} - (\frac{1}{21}) + \frac{1}{20} - (\frac{1}{22}) ]\)
Sum of 20 terms \(= (\frac{1}{2})[ (\frac{1}{1}) + (\frac{1}{2}) - (\frac{1}{21}) - (\frac{1}{22}) ]\)
Sum of 20 terms = \(\frac{325}{462}\)
Answer: Option B
27 Feb 2019, 05:41
Kudos
Bookmarks
Solution
Given
- • The nth term of a sequence a1, a2, a3, … an is given by an = \(\frac{1}{n(n+2)}\)
To Find
- • The sum of the first 20 terms of the sequence.
Approach & Working
We are given an = \(\frac{1}{n(n+2)}\).
- • Hence, a1 = \(\frac{1}{1(1+2)} =\frac{1}{(1×3)}\)
• Similarly, a2 = \(\frac{1}{(2(2+2))}=\frac{1}{(2×4)}\)
- o a3 = \(\frac{1}{(3×5)}\)
o a4 =\(\frac{1}{(4×6)}\) and so on.
o a19 = \(\frac{1}{(19×21)}\)
o a20 = \(\frac{1}{(20×22)}\)
- • Hence, a1 + a2 + a3 + … +a19 + a20 =\(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}\)
Let us assume a1 + a2 + a3 + … a20 = S.
- • S = \(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}\)
As the denominator of each fraction is a product of 2 numbers with a constant difference of 2, let us multiply both side by 2 so that we can express the numerator of each fraction in terms of its denominator
Let us multiply by 2 on both the sides.
- • 2S = 2(\(\frac{1}{(1×3)}\) + \(\frac{1}{(2×4)}\) + \(\frac{1}{(3×5)}\) + \(\frac{1}{(4×6)}\) + ... + \(\frac{1}{(19×21)}\) + \(\frac{1}{(20×22)}) 2\)
• 2S = \(\frac{2}{(1×3)}\) + \(\frac{2}{(2×4)}\) + \(\frac{2}{(3×5)}\) + \(\frac{2}{(4×6)}\) + ... + \(\frac{2}{(19×21)}\) + \(\frac{2}{(20×22)}\)
- o If we observe carefully then,
- • \(\frac{2}{(1×3)} = \frac{1}{1}- \frac{1}{3}\)
• \(\frac{2}{(2×4)}= \frac{1}{2}- \frac{1}{4}\)
• \(\frac{2}{(3×5)} = \frac{1}{3} - \frac{1}{5}\)
• \(\frac{2}{(4×6)} = \frac{1}{4} - \frac{1}{6}\)
• Similarly, \(\frac{2}{(19×21)} = \frac{1}{19}- \frac{1}{21}\) and \(\frac{2}{(20×22)} = \frac{1}{20} - \frac{1}{22}\)
- o Now, we can observe that odd or even numbered terms are beginning with the same number that the preceding number is ending with and their sign is opposite.
o So, they will get cancelled when added.
- • For example: \(\frac{2}{(1×3)} = \frac{1}{1} - \frac{1}{3} and \frac{2}{(3×5)} = \frac{1}{3} - \frac{1}{5}\) (odd numbered terms)
- o 1/3 will get cancelled when these two terms are added.
- • And, we will get 1- 1/5 i.e. First number of first term - last number of last term.
o Thus, if we add all the odd terms till a20 then:
- • Sum = First number of first odd term - last number of last odd term.
- • \(\frac{2}{(2×4)} = \frac{1}{2} - \frac{1}{4}\) and \(\frac{2}{(4×6)} = \frac{1}{4} - \frac{1}{6}\) (Even numbered terms)
- o As we did with odd terms, the sum of all the even terms till a20 = First number of first even term - last number of last even term.
Thus, 2S = 2a1 + 2a2 + 2a3 + ... + 2a19+ 2a20
- • Or, 2S = Sum of all odd terms till a20 + Sum of all even terms till a20
- o = (2a1 + 2a3 + ... + 2a19) + (2a2 + 2a4 + ... + 2a20)
o = \([ (\frac{1}{1} - \frac{1}{3})+ ... +(\frac{1}{19} - \frac{1}{21}) ] + [ (\frac{1}{2} - \frac{1}{4})+ … +(\frac{1}{20} - \frac{1}{22}) ]\)
o = \((\frac{1}{1} - \frac{1}{21}) + (\frac{1}{2} - \frac{1}{22})\)
o 2S = \(\frac{20}{21} + \frac{10}{22}\)
o S = \(\frac{10}{21}+ \frac{5}{22}\)
o S = \(\frac{325}{462}\)
Hence, option B is the correct answer.
Correct Answer: Option B
