December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. December 16, 2018 December 16, 2018 03:00 PM EST 04:00 PM EST Strategies and techniques for approaching featured GMAT topics
Author 
Message 
TAGS:

Hide Tags

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3326
Location: India
GPA: 3.12

In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
28 Mar 2018, 09:17
Question Stats:
64% (03:17) correct 36% (03:00) wrong based on 52 sessions
HideShow timer Statistics
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=? A. 7 B. 9 C. 10 D. 12 E. 15 Source: Experts Global
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
You've got what it takes, but it will take everything you've got



PS Forum Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 425

Re: In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
28 Mar 2018, 11:52
pushpitkc wrote: In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=? A. 7 B. 9 C. 10 D. 12 E. 15 Source: Experts Global The first m terms of the sequence form an increasing AP with A as 50 & d = 5
So mth term is given by A + (m  1)*d ( formula for nth term of AP) this is 50 +(m1)5.There is a second AP with the mth term of prev being it's first term hence A & d = 2.We are given the last term of total sequence is also 50.... so we need to find the number of terms in the second AP. mth term is 1st term, and we have 29  m terms (since there are 29 terms in all)after that... hence 30  m terms in all in second AP. Hence the last term is the (30  m)th term of 2nd AP. Applying the formula we get.... 50 = A' + (n'  1)d' .... A' is mth term from above 1st AP, d' = 2 and n' is (30  m) Plugging in the values: \(50 = 50 +(m1)5 + (30  m  1)*2\) \((29  m)*2 = 5m  5\) ....( 50's cancel and 2 term goes to the LHS) \(58 + 5 = 7m\) \(63 = 7m\) \(m = 9\) Hence Option (B) is correct. Best, Gladi



VP
Joined: 07 Dec 2014
Posts: 1129

In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
Updated on: 25 Apr 2018, 14:33
pushpitkc wrote: In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=? A. 7 B. 9 C. 10 D. 12 E. 15 Source: Experts Global50+(m1)*5=50+(29m)*2 m=9 B
Originally posted by gracie on 28 Mar 2018, 16:54.
Last edited by gracie on 25 Apr 2018, 14:33, edited 1 time in total.



Intern
Joined: 25 Sep 2017
Posts: 3

Re: In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
23 Apr 2018, 23:09
pushpitkc wrote: In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=? A. 7 B. 9 C. 10 D. 12 E. 15 Source: Experts Global Bunuel Hi Bunuel, can you please explain me this problem? Thanks, Nikita



PS Forum Moderator
Status: It always seems impossible until it's done.
Joined: 16 Sep 2016
Posts: 425

Re: In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
23 Apr 2018, 23:39
NikitaS wrote: Bunuel Hi Bunuel, can you please explain me this problem? Thanks, Nikita Hi NikitaS, Can you please specify which part is troubling you? I have provided with a detailed solution as follows... Kindly go through it once and let me know. I would be happy to help Best, Gladi Gladiator59 wrote: pushpitkc wrote: In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=? A. 7 B. 9 C. 10 D. 12 E. 15 Source: Experts Global The first m terms of the sequence form an increasing AP with A as 50 & d = 5
So mth term is given by A + (m  1)*d ( formula for nth term of AP) this is 50 +(m1)5.There is a second AP with the mth term of prev being it's first term hence A & d = 2.We are given the last term of total sequence is also 50.... so we need to find the number of terms in the second AP. mth term is 1st term, and we have 29  m terms (since there are 29 terms in all)after that... hence 30  m terms in all in second AP. Hence the last term is the (30  m)th term of 2nd AP. Applying the formula we get.... 50 = A' + (n'  1)d' .... A' is mth term from above 1st AP, d' = 2 and n' is (30  m) Plugging in the values: \(50 = 50 +(m1)5 + (30  m  1)*2\) \((29  m)*2 = 5m  5\) ....( 50's cancel and 2 term goes to the LHS) \(58 + 5 = 7m\) \(63 = 7m\) \(m = 9\) Hence Option (B) is correct. Best, Gladi



Math Expert
Joined: 02 Sep 2009
Posts: 51223

Re: In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
23 Apr 2018, 23:46
pushpitkc wrote: In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=? A. 7 B. 9 C. 10 D. 12 E. 15 Source: Experts GlobalFirst, we have m integers, for which each term is 5 greater than the previous integer: 50, 55, 60, 65, 70, ..., 50 + 5(m  1) Notice, that the last integer in this, first, part of the sequence is \(50 + 5(m  1)\). \(n_{th}\) term in the arithmetic progression (AP) is given by \(a_n=a_1+d(n1)\), where d is the common difference. After those m integers, each integer is 2 less than the previous integer: \([50 + 5(m  1)]  2\), \([50 + 5(m  1)]  4\), \([50 + 5(m  1)]  6\), ..., 50. Notice, that since there are a total of 29 integers in the whole sequence, then the above continues for 29  m integers. The last integer given to be 50, thus by applying the same rule for AP, we'll get \(50 = [50 + 5(m  1)  2] + (2)[(29  m)  1]\): \(0 = 5m  7 56 + 2m\) \(7m=63\); \(m=9\). Answer: B. 12. Sequences For other subjects check: ALL YOU NEED FOR QUANT. Ultimate GMAT Quantitative MegathreadHope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 25 Sep 2017
Posts: 3

Re: In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
24 Apr 2018, 23:08
Hi Gladiator59 , Thank you for your reply...I missed to see your explanation!! Thanks for your help.



Intern
Joined: 25 Sep 2017
Posts: 3

Re: In a sequence of integers, each term in the first m integers is 5
[#permalink]
Show Tags
24 Apr 2018, 23:09
Thanks Bunuel




Re: In a sequence of integers, each term in the first m integers is 5 &nbs
[#permalink]
24 Apr 2018, 23:09






