Author 
Message 
Intern
Joined: 22 Jan 2009
Posts: 1

2
This post was BOOKMARKED
A computer generated a sequence \(A\) of numbers using the following formula: \(A_n = A_1 + (n1)d\) \(d\) is the common difference between any two consecutive terms of the sequence \(A\) If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence? (A) 3 (B) 2 (C) 1 (D) 1 (E) 3 Source: GMAT Club Tests  hardest GMAT questions Can someone explain this? I am not quite sure how the above notations transform... Thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 39704

Re: M03#29 [#permalink]
Show Tags
03 Jan 2011, 04:07
gmatdelhi wrote: tingle15 wrote: I had the same issue. Because it said consecutive, I assumed d = 1. Are we wrong in some way? When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x3, x2, x1, x, x+1, x+2, .... For example: 7, 6, 5 are consecutive integers. 2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers. 3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers. But in the original question, stem says consecutive terms of the sequence A (which is given to be arithmetic progression) and not consecutive integers, so the common difference is not necessary to be 1 in this case. As for the question: A computer generated a sequence \(A\) of numbers using the following formula: \(A_n = A_1 + (n1)d\). \(d\) is the common difference between any two consecutive terms of the sequence \(A\). If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence? A. 3 B. 2 C. 1 D. 1 E. 3 Given: \(a_2+a_5=(a_1+d)+(a_1+4d)=2a_1+5d=8\); \(a_3+a_7=(a_1+2d)+(a_1+6d)=2a_1+8d=14\); Subtract 1 from 2: \(3d=6\) > \(d=2\) > as \(2a_1+5d=8\) then \(2a_1+5*2=8\) > \(a_1=1\). Answer: D. Hope it helps.
_________________
New to the Math Forum? Please read this: 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



Director
Joined: 04 Jan 2008
Posts: 898

Re: M03#29 [#permalink]
Show Tags
14 Mar 2009, 21:48
1
This post received KUDOS
its AP A, A+d,A+2d,A+3d,.........,A+(n1)d sorry I cant able to write the equation here but its a simple problem solve 2 equations for 2 unknowns(A and d) ThinkingHat wrote: Quote: A computer generated a sequence \(A\) of numbers using the following formula: \(A_n = A_1 + (n1)d\)
\(d\) is the common difference between any two consecutive terms of the sequence \(A\)
If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence?
(C) 2008 GMAT Club  m03#29
* 3 * 2 * 1 * 1 * 3
\(\left{ \begin{eqnarray*} A_2+A_5 &=& A_1 + d + A_1 + 4d = 2A_1 + 5d = 8\\ A_3+A_7 &=& A_1 + 2d + A_1 + 6d = 2A_1 + 8d = 14\\ \end{eqnarray*}\) \(\left{ \begin{eqnarray*} d &=& 2\\ A_1 &=& 1\\ \end{eqnarray*}\)
The correct answer is D.
Can someone explain this? I am not quite sure how the above notations transform... Thanks!
_________________
http://gmatclub.com/forum/mathpolygons87336.html http://gmatclub.com/forum/competitionforthebestgmaterrorlogtemplate86232.html



Senior Manager
Joined: 13 Dec 2009
Posts: 262

Re: M03#29 [#permalink]
Show Tags
22 May 2010, 08:24
1
This post received KUDOS
The wording of the question is rather confusing. When I did this question the wording was: A computer generated a consecutive set of numbers A using the following formula: Can someone remove the word consecutive from the question? It totally misled me into thinking that the set of numbers are consecutive integers like 1,2,3.. The question should be phrased like below: A computer generated a consecutive set of numbers \(A\)using the following formula: \(A n = A 1 + (n1)d\) where \(d\)is the common difference between any number and the next number in the set \(A\).
_________________
My debrief: doneanddusted730q49v40



Manager
Joined: 07 Jul 2007
Posts: 137

Re: M03#29 [#permalink]
Show Tags
16 Mar 2009, 08:51
nitya34 wrote: its AP A, A+d,A+2d,A+3d,.........,A+(n1)d sorry I cant able to write the equation here but its a simple problem solve 2 equations for 2 unknowns(A and d) ThinkingHat wrote: Quote: A computer generated a sequence \(A\) of numbers using the following formula: \(A_n = A_1 + (n1)d\)
\(d\) is the common difference between any two consecutive terms of the sequence \(A\)
If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence?
(C) 2008 GMAT Club  m03#29
* 3 * 2 * 1 * 1 * 3
\(\left{ \begin{eqnarray*} A_2+A_5 &=& A_1 + d + A_1 + 4d = 2A_1 + 5d = 8\\ A_3+A_7 &=& A_1 + 2d + A_1 + 6d = 2A_1 + 8d = 14\\ \end{eqnarray*}\) \(\left{ \begin{eqnarray*} d &=& 2\\ A_1 &=& 1\\ \end{eqnarray*}\)
The correct answer is D.
Can someone explain this? I am not quite sure how the above notations transform... Thanks! Response:  It is an arithmetic progression sequence where you have formula: \(a_n = a_1 + (n1)d\) And now if you put n = 5, n = 2, n =7 you will get these forumuas.



Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: M03#29 [#permalink]
Show Tags
30 Sep 2010, 09:28
2a+5d=8 and 2a+8d=14 => d=3 and a = 1. Answer D wins.
_________________
Support GMAT Club by putting a GMAT Club badge on your blog



Manager
Joined: 08 Sep 2010
Posts: 226
Location: India
WE 1: 6 Year, Telecom(GSM)

Re: M03#29 [#permalink]
Show Tags
30 Sep 2010, 10:04
An=A1(n1)d is the formula for nth term in an arithmetic progression. hence 7th term A7=A1(71)d Similarly,2nd term A2=A1(21)d 3rd term A3=A1(31)d and 5th term A5=A1(51)d Now According to question,A2+A5=8 i.e 2A1+5d=8 and A3+A7=14 i.e, 2A1+8d=14 Solving these two we will get A1=1 Hence the answer is D
_________________
Consider KUDOS if You find it good



Intern
Joined: 19 Jul 2010
Posts: 5

Re: M03#29 [#permalink]
Show Tags
30 Sep 2010, 16:12
Arithmetic sequence.



Intern
Joined: 18 Aug 2010
Posts: 29
Concentration: Entrepreneurship, Finance
GPA: 3.6
WE: Engineering (Telecommunications)

Re: M03#29 [#permalink]
Show Tags
30 Sep 2010, 17:04
D.
I tried backsubstitution and didn't get anywhere. had to solve it the oldfashioned  2 equations, 2 variables way.



Manager
Joined: 15 Apr 2010
Posts: 167

Re: M03#29 [#permalink]
Show Tags
12 Oct 2010, 13:27
sidhu4u wrote: The wording of the question is rather confusing. When I did this question the wording was:
A computer generated a consecutive set of numbers A using the following formula:
Can someone remove the word consecutive from the question? It totally misled me into thinking that the set of numbers are consecutive integers like 1,2,3..
The question should be phrased like below:
A computer generated a consecutive set of numbers \(A\)using the following formula:
\(An = A1 + (n1)d\) where \(d\)is the common difference between any number and the next number in the set \(A\). I agree... Can someone please provide a resolution for this...



Manager
Joined: 23 Oct 2009
Posts: 84
Location: New Delhi, India
Schools: Chicago Booth, Harvard, LBS, INSEAD, Columbia

Re: M03#29 [#permalink]
Show Tags
03 Jan 2011, 01:16
tingle15 wrote: sidhu4u wrote: The wording of the question is rather confusing. When I did this question the wording was:
A computer generated a consecutive set of numbers A using the following formula:
Can someone remove the word consecutive from the question? It totally misled me into thinking that the set of numbers are consecutive integers like 1,2,3..
The question should be phrased like below:
A computer generated a consecutive set of numbers \(A\)using the following formula:
\(An = A1 + (n1)d\) where \(d\)is the common difference between any number and the next number in the set \(A\). I agree... Can someone please provide a resolution for this... I had the same issue. Because it said consecutive, I assumed d = 1. Are we wrong in some way?
_________________
Read about my GMAT prep at http://gmatting.blogspot.com/ 1st Feb '11  Actual GMAT : 730 (Q48 V42) AWA 6.0
My Practice GMAT Scores 29th Jan '11  GMATPrep#2 : 700 (Q47 V38) 23rd Jan '11  MGMAT Practice Test #3 : 670 (Q45 V36) 19th Jan '11  GMATPrep#1 v.1 : 710 (Q49 V37) 15th Jan '11  GMATPrep#1 : 720 (Q47 V42) 11th Jan '11  MGMAT Practice Test #2 : 740 (Q47 V44) 6th Jan '11  Kaplan#2 : 620 (Q40 V35) 28th Dec '10  PowerPrep#1 : 670 (Q47 V35) 30th Oct '10  MGMAT Practice Test #1 : 660 (Q45 V35) 12th Sept '10  Kaplan Free Test : 610 (Q39 V37) 6th Dec '09  PR CAT #1 : 650 (Q44 V37) 25th Oct '09  GMATPrep#1 : 620 (Q44 V34)
If you feel like you're under control, you're just not going fast enough. A goal without a plan is just a wish. You can go higher, you can go deeper, there are no boundaries above or beneath you.



Manager
Joined: 20 Nov 2010
Posts: 218

Re: M03#29 [#permalink]
Show Tags
04 Oct 2011, 11:02
Easy one. Its an AP with d as common difference and A1 as first term.
_________________
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ MGMAT 6 650 (51,31) on 31/8/11 MGMAT 1 670 (48,33) on 04/9/11 MGMAT 2 670 (47,34) on 07/9/11 MGMAT 3 680 (47,35) on 18/9/11 GMAT Prep1 680 ( 50, 31) on 10/11/11
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ CR notes http://gmatclub.com/forum/massivecollectionofverbalquestionsscrcandcr106195.html#p832142 http://gmatclub.com/forum/1001dsquestionsfile106193.html#p832133 http://gmatclub.com/forum/gmatprepcriticalreasoningcollection106783.html http://gmatclub.com/forum/howtoget60awamyguide64327.html http://gmatclub.com/forum/howtoget60awamyguide64327.html?hilit=chineseburned



Manager
Joined: 11 Feb 2011
Posts: 74

Re: M03#29 [#permalink]
Show Tags
04 Oct 2011, 17:21
easy one..D is the answer



Manager
Joined: 16 Sep 2010
Posts: 220
Location: United States
Concentration: Finance, Real Estate

Re: M03#29 [#permalink]
Show Tags
14 Nov 2011, 23:19
ThinkingHat wrote: A computer generated a sequence \(A\) of numbers using the following formula: \(A_n = A_1 + (n1)d\) \(d\) is the common difference between any two consecutive terms of the sequence \(A\) If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence? (A) 3 (B) 2 (C) 1 (D) 1 (E) 3 Source: GMAT Club Tests  hardest GMAT questions Can someone explain this? I am not quite sure how the above notations transform... Thanks! Its pretty easy when you set it up as equations to solve. Equation one is = A + (21)d + A + (51)d = 8 => 2A + 5d = 8 Equation two is = A + (31)d + A + (71)d = 14 => 2A + 8d = 14 Since the only difference between the two equations is 3d = 6 we know that d = 2. Plug that into one of the equations and you get A = 1.



Senior Manager
Joined: 23 Oct 2010
Posts: 383
Location: Azerbaijan
Concentration: Finance

Re: M03#29 [#permalink]
Show Tags
17 Dec 2011, 04:34
a2+a5=8 a3+a7=14 or a2+d+a6+d=14 a2+24+a5+d=14 8+3d=14 d=2 a1+d+a1+4d=8 2a1+5d=8 2a1=5*2=8 a1=1
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 537
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)

Re: M03#29 [#permalink]
Show Tags
04 Oct 2012, 12:04
i don't know how long the GMAC will continue to ask such question in the real test because this one is easy yet time consuming. A2+A5 => { A1+(21)d}+{A1+(51)d=8 => 2A1+5d=8(i) A3+A7 => {A1+(31)d}+{A1+(71)d=14 => 2A1+8d=14(ii) solve (i) & (ii) d=2 putting the value of d in (i) [or (ii)] A1=1 D wins
_________________
" Make more efforts " Press Kudos if you liked my post










