Last visit was: 21 Nov 2024, 14:15 It is currently 21 Nov 2024, 14:15

Close

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

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

Track
Your Progress

every week, we’ll send you an estimated GRE 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.

Close

Request Expert Reply

Confirm Cancel
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4812
Own Kudos [?]: 11188 [3]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4812
Own Kudos [?]: 11188 [3]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
avatar
Intern
Intern
Joined: 03 Jan 2019
Posts: 3
Own Kudos [?]: 4 [0]
Given Kudos: 0
Send PM
avatar
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [3]
Given Kudos: 0
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
3
Expert Reply
sandy wrote:
The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?

(A) 243
(B) 14,400
(C) 14,500
(D) 24,300
(E) 24,545



Another method..

This is an AP as terms are evenly spaced..
\(A_1=45\), so \(A_{100}=45+2(100-1)=45+198=243\)
the average of the sequence is \(\frac{1^{st} term+ 2^{nd}}{2}=\frac{45+243}{2}=\frac{288}{2}=144\)
SUM = average * number of terms = 144*100 = 14,400

B
avatar
Intern
Intern
Joined: 26 Dec 2018
Posts: 2
Own Kudos [?]: 4 [0]
Given Kudos: 0
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
3
sandy wrote:
The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?

(A) 243
(B) 14,400
(C) 14,500
(D) 24,300
(E) 24,545


Solution:

Given A_1 = 45
n ≥ 2,
A_2 = A_1 + 2 = 45+2 = 47

Series {45, 47, 49.....)
First term a = 45 and common difference d = 2
Number of terms = 100

Sum = (n/2)*(2a + (n-1)d)
= (100/2) * (2 x 45 + 99 x 2)
= (100/2) * 288
= 14400
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 711 [0]
Given Kudos: 161
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
chetan2u wrote:
sandy wrote:
The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?

(A) 243
(B) 14,400
(C) 14,500
(D) 24,300
(E) 24,545



Another method..

This is an AP as terms are evenly spaced..
\(A_1=45\), so \(A_{100}=45+2(100-1)=45+198=243\)
the average of the sequence is \(\frac{1^{st} term+ 2^{nd}}{2}=\frac{45+243}{2}=\frac{288}{2}=144\)
SUM = average * number of terms = 144*100 = 14,400

B



the average of the sequence is \(\frac{1^{st} term+ 2^{nd}}{2}=\frac{45+243}{2}=\frac{288}{2}=144\)
Here it,s not 2nd term(the 243) it will be last term, edit this, Thanks
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 711 [0]
Given Kudos: 161
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
sbingi wrote:
sandy wrote:
The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?

(A) 243
(B) 14,400
(C) 14,500
(D) 24,300
(E) 24,545


Solution:

Given A_1 = 45
n ≥ 2,
A_2 = A_1 + 2 = 45+2 = 47

Series {45, 47, 49.....)
First term a = 45 and common difference d = 2
Number of terms = 100

Sum = (n/2)*(2a + (n-1)d)
= (100/2) * (2 x 45 + 99 x 2)
= (100/2) * 288
= 14400


Avoid as much as possible the previous rules that we taught to memorize in School :) :) :)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12196 [0]
Given Kudos: 136
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
2
sandy wrote:
The sequence A is defined by \(A_n = A_{n – 1} + 2\) for each integer n ≥ 2, and \(A_1 = 45\). What is the sum of the first 100 terms in sequence A?

(A) 243
(B) 14,400
(C) 14,500
(D) 24,300
(E) 24,545


Let's examine a few terms to see the pattern:
term1 = 45 (we add 0 two's)
term2 = 45 + 2 = 47 (we add 1 two)
term3 = 45 + 2 + 2 = 49 (we add 2 two's)
term4 = 45 + 2 + 2 + 2 = 51 (we add 3 two's)
.
.
.
term100 = 45 + 2 + 2 ....... + 2 = 243 (we add 99 two's)

So, the sum of the first 100 terms = 45 + 47 + 49 + . . . + 241 + 243
Let's add the values in PAIRS, by pairing up values from each side (left and right) of the sum.

That is: 45 + 47 + 49 + . . . + 239 + 241 + 243 = (45 + 243) + (47 + 241) + (49 + 239) + ....
= (288) + (288) + (288) + ....
Since we have 50 PAIRS that each add to 288, the TOTAL sum = (50)(288) = 14,400

Answer: B

Cheers,
Brent
avatar
Manager
Manager
Joined: 22 May 2019
Posts: 58
Own Kudos [?]: 51 [0]
Given Kudos: 0
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
sandy wrote:
Explanation

The first term of the sequence is 45, and each subsequent term is determined by adding 2. The problem asks for the sum of the first 100 terms, which cannot be calculated directly in the given time frame; instead, find the pattern.

The first few terms of the sequence are 45, 47, 49, 51,… What’s the pattern? To get to the 2nd term, start with 45 and add 2 once. To get to the 3rd term, start with 45 and add 2 twice. To get to the 100th term, then, start with 45 and add 2 ninety-nine times:

\(45 + (2)(99) = 243.\)

Next, find the sum of all odd integers from 45 to 243, inclusive. To sum up any evenly spaced set, multiply the average (arithmetic mean) by the number of elements in the set. To get the average, average the first and last terms. Since \(\frac{45+243}{2}= 144\), the average is 144.


To find the total number of elements in the set, subtract 243 – 45 = 198, then divide by 2 (count only the odd numbers, not the even ones): \(\frac{198}{2}= 99\) terms.

Now, add 1 (to count both endpoints in a consecutive set, first subtract and then “add 1 before you’re done”). The list has 100 terms. Multiply the average and the number of terms:

\(144 \times 100 = 14,400\)

It has been explicitly said that the sum of the first 100 terms. Isn't it enough to move on instead of calculating the total number of terms in the above way?
User avatar
Manager
Manager
Joined: 19 Nov 2018
Posts: 102
Own Kudos [?]: 158 [0]
Given Kudos: 0
Send PM
Re: The sequence A is defined by An = An – 1 + 2 for each intege [#permalink]
Here is the formula for finding a specific term in an arithmetic sequence. I didn't do the whole problem, because everyone else already has. I just thought this part might have been glossed over a little.
Attachments

daum_equation_1566334680442.png
daum_equation_1566334680442.png [ 98.83 KiB | Viewed 11294 times ]

User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5030
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: The sequence A is defined by An = An 1 + 2 for each intege [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
Re: The sequence A is defined by An = An 1 + 2 for each intege [#permalink]
Moderators:
GRE Instructor
84 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne