GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 23 Jan 2020, 22:52 ### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Question of the Week - 37 (The nth term of a sequence...)

Author Message
TAGS:

### Hide Tags

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3219
Question of the Week - 37 (The nth term of a sequence...)  [#permalink]

### Show Tags

1
6 00:00

Difficulty:   95% (hard)

Question Stats: 28% (02:44) correct 73% (02:47) wrong based on 40 sessions

### HideShow timer Statistics

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. $$\frac{650}{462}$$

_________________

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.
CEO  D
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 3026
Location: India
GMAT: INSIGHT
Schools: Darden '21
WE: Education (Education)
Re: Question of the Week - 37 (The nth term of a sequence...)  [#permalink]

### Show Tags

EgmatQuantExpert wrote:
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. $$\frac{650}{462}$$

$$a_n = \frac{1}{n(n+2)}$$

$$a_1 = \frac{1}{1(1+2)} = \frac{1}{1*3} = \frac{1}{2}[(\frac{1}{1}) - (\frac{1}{3})]$$
$$a_2 = \frac{1}{2(2+2)} = \frac{1}{2*4}= \frac{1}{2}[(\frac{1}{2}) - (\frac{1}{4})]$$
$$a_3 = \frac{1}{3(3+2)} = \frac{1}{3*5}= \frac{1}{2}[(\frac{1}{3}) - (\frac{1}{5})]$$
$$a_4 = \frac{1}{4(4+2)} = \frac{1}{4*6}= \frac{1}{2}[(\frac{1}{4}) - (\frac{1}{6})]$$
$$a_5 = \frac{1}{5(5+2)} = \frac{1}{5*7}= \frac{1}{2}[(\frac{1}{5}) - (\frac{1}{7})]$$

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 = $$\frac{325}{462}$$

_________________
Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

ACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3219
Re: Question of the Week - 37 (The nth term of a sequence...)  [#permalink]

### Show Tags

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 Similarly, if we add the next odd term then 1/3 will get cancelled.
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.

_________________ Re: Question of the Week - 37 (The nth term of a sequence...)   [#permalink] 27 Feb 2019, 05:41
Display posts from previous: Sort by

# Question of the Week - 37 (The nth term of a sequence...)  