GRE Prep 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 25Crafting an Impactful Resume: Balancing Detail with Brevity
10:00 AM EDT
-10:30 AM EDT
Showcase your achievements through impactful storytelling! Get ready for an LIVE session on April 25, Learn how to craft compelling narratives on your resume that resonate with admission committees.
Apr 26Get a Top GRE Score with Target Test Prep
12:00 PM EDT
-01:00 PM EDT
The Target Test Prep course represents a quantum leap forward in GRE preparation, a radical reinterpretation of the way that students should study. In fact, more and more Target Test Prep students are scoring 329 and above on the GRE.
In the sequence a1, a2, a3,....a100, the kth term is defined
[#permalink]
24 Feb 2016, 15:57
24 Feb 2016, 15:57
4
8
Bookmarks
00:00
Question Stats:
81% (03:23) correct
18% (02:00) wrong based on 53 sessions
Hide Show timer Statistics
In the sequence a1, a2, a3,....a100, the kth term is defined as \(ak= \frac{1}{k} - \frac{1}{k+1}\) for all integers k from 1 through 100. What is the sum of 100 terms of the sequence?
A. \(\frac{1}{10100}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{100}{101}\)
E. \(1\)
A. \(\frac{1}{10100}\)
B. \(\frac{1}{100}\)
C. \(\frac{1}{101}\)
D. \(\frac{100}{101}\)
E. \(1\)
Director
Joined: 16 May 2014
Posts: 592
Given Kudos: 0
GRE 1: Q165 V161
Re: In the sequence a1, a2, a3,....a100, the kth term is defined
[#permalink]
25 Feb 2016, 04:24
25 Feb 2016, 04:24
14
1
Expert Reply
2
Bookmarks
Answer :
D \(\frac{100}{101}\)
Given
a_k = \(\frac{1}{k}\)-\(\frac{1}{k+1}\);
So the sum of a_1 to a_100 is explained below-
a_1 + a_2 + a_3 + ...... + a_100
= \(\frac{1}{1}\) - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\)+ ..... + \(\frac{1}{100}\) - \(\frac{1}{101}\)
= 1- \(\frac{1}{101}\)
= \(\frac{100}{101}\)
D \(\frac{100}{101}\)
Explanation:
Given
a_k = \(\frac{1}{k}\)-\(\frac{1}{k+1}\);
So the sum of a_1 to a_100 is explained below-
a_1 + a_2 + a_3 + ...... + a_100
= \(\frac{1}{1}\) - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\)+ ..... + \(\frac{1}{100}\) - \(\frac{1}{101}\)
= 1- \(\frac{1}{101}\)
= \(\frac{100}{101}\)
General Discussion
Re: In the sequence a1, a2, a3,....a100, the kth term is defined
[#permalink]
29 Jan 2019, 00:27
29 Jan 2019, 00:27
3
hello, I do not understand from the point of 1-1/101. How do you come to a conclusion that the sum is 1-1/101?
