GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Aug 2018, 18:27

### 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

# If the first, third and thirteenth terms of an arithmetic

Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Feb 2012
Posts: 47
If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

18 Aug 2012, 13:39
2
7
00:00

Difficulty:

55% (hard)

Question Stats:

74% (02:28) correct 26% (02:30) wrong based on 180 sessions

### HideShow timer Statistics

If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?

A. 2
B. 3
C. 4
D. 5
E. 6
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8187
Location: Pune, India
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

30 Aug 2012, 22:37
13
2
NYC5648 wrote:
If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?

A. 2
B. 3
C. 4
D. 5
E. 6

The following was my thought process:

It looks like a simple AP of small integers.

I will start with some concrete info. I know that 'sum of the fourth and seventh terms of this arithmetic progression is 40'
(a+3d) + (a + 6d) = 40
2a + 9d = 40

Note here that 2a is even and sum of 2a and 9d is even. So, 9d must be even too. This means, d, the common difference, must be even.
If d = 2, 9d = 18 and 2a = 40 - 18 = 22. But a = 11 is not in the options so ignore it.
If d = 4, 9d = 36 and 2a = 40 - 36 = 4. This gives us a = 2

Now check: a = 2, d = 4
1st term = 2, 3rd term = 10, 13th term = 50.
2, 10 and 50 are in GP. Good!
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### General Discussion Director Joined: 22 Mar 2011 Posts: 604 WE: Science (Education) Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 18 Aug 2012, 14:20 1 NYC5648 wrote: If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence? a) 2 b) 3 c) 4 d) 5 e) 6 Answer: a) Thanks! We can write $$a_1*a_{13}=a_3*a_3$$ and $$a_3+a_7=40.$$ Since $$a_3=a_1+2d,$$ (where $$d$$ is the the constant defining the arithmetic progression), $$a_{13}=a_1+12d$$, $$\, a_4=a_1+3d$$, $$\, a_7=a_1+6d$$, from the above two equations we get $$a_1(a_1+12d)=(a_1+2d)^2$$ and $$a_1+3d+a_1+6d=40.$$ Further, $$12a_1d=4a_1d+4d^2$$ or $$8a_1d=4d^2$$ which leads to $$2a_1=d.$$ (If $$d=0$$ all the terms in the arithmetic progression are equal to 20, and so are those in the geometric progression. It should have been stated that the arithmetic progression is non-constant.) The other equation leads to $$2a_1+9d=40$$, so $$2a_1+18a_1=40,$$ or $$a_1=2.$$ Answer A. IMO, not a real GMAT question. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Intern Joined: 05 Jul 2011 Posts: 1 Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 30 Aug 2012, 00:57 In my opinion one can solve this question by simple calculations pick an option from choices lets say 2 first one is arith matic sequence so it will be like 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 notice that square of two is 4 and square of 4 is i6 which is the thirteenth term so one can solve it by mental math. Intern Joined: 28 Aug 2012 Posts: 44 Location: Austria GMAT 1: 770 Q51 V42 Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 30 Aug 2012, 04:25 1 shoaibfu wrote: In my opinion one can solve this question by simple calculations pick an option from choices lets say 2 first one is arith matic sequence so it will be like 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 notice that square of two is 4 and square of 4 is i6 which is the thirteenth term so one can solve it by mental math. That's wrong. The solution is: 2, 6, 10, 14, 18, 22, 26, ..., 50. Geometric means constant factor, so in this case: 2, 10, 50 (constant factor being 5). And 14+26=40. It would be pretty hard to guess the correct solution. A lot of trial and error. It would take too long. In your sequence, 16 isn't even the 13th term. It's the 15th. Intern Joined: 12 Jul 2012 Posts: 4 Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 30 Aug 2012, 07:26 2 If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence? if 1 ,3 and 13 in ap are gp then (a+2d)sqr = a(a+12a) and a+3d + a+6d = 40 solve the two eq.. we get a = 2 Manager Status: exam is close ... dont know if i ll hit that number Joined: 06 Jun 2011 Posts: 167 Location: India Concentration: International Business, Marketing GMAT Date: 10-09-2012 GPA: 3.2 Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 30 Aug 2012, 20:35 Zinsch123 wrote: shoaibfu wrote: In my opinion one can solve this question by simple calculations pick an option from choices lets say 2 first one is arith matic sequence so it will be like 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 notice that square of two is 4 and square of 4 is i6 which is the thirteenth term so one can solve it by mental math. That's wrong. The solution is: 2, 6, 10, 14, 18, 22, 26, ..., 50. Geometric means constant factor, so in this case: 2, 10, 50 (constant factor being 5). And 14+26=40. It would be pretty hard to guess the correct solution. A lot of trial and error. It would take too long. In your sequence, 16 isn't even the 13th term. It's the 15th. exactly 16 will be 15th term not 13th that itself can refute further calculation based on that approach _________________ just one more month for exam... Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8187 Location: Pune, India Re: If the first, third and thirteenth terms of an arithmetic [#permalink] ### Show Tags 30 Aug 2012, 22:46 metallicafan wrote: +1 A Although it requires a lot of trial and error to find how to combine the equalities. :s Probably, a 700+ question. Zinsch123 wrote: It would be pretty hard to guess the correct solution. A lot of trial and error. It would take too long. In your sequence, 16 isn't even the 13th term. It's the 15th. Actually, it isn't very hard to guess. When you use 'brute force'/'hit and trial'/'logic'/'intuition', you need to have a point to start from. You can certainly not say, "Ok, say the first term is 2 and common difference is 1 - not working. Say, common difference is 2 - not working etc etc." You first need to analyze whatever info you have and shrink your world of possible APs. Otherwise, using the theoretical algebraic solution might actually be faster. The good thing is that as far as GMAT is concerned, a combination of logic and brute force works very well. Often, you will get the answer using logic alone. Sometimes, a couple of hit and trials are needed too. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

MBA Section Director
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 5251
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

24 Sep 2013, 06:56
2
If the first, third and thirteenth terms of an arithmetic progression are in geometric progression, and the sum of the fourth and seventh terms of this arithmetic progression is 40, find the first term of the sequence?

Slightly different algebraic approach.........

the first, third and thirteenth terms of an arithmetic progression are in geometric progression

First term = a
Third Term = a+2d
Thirteenth Term = a+12d

They are in Geometric Progression too
So $$\frac{(a+2d)}{a} = \frac{(a+12d)}{(a+2d)}$$ --------------------------> {When a, b, c, d are in GP then b/a = c/b = d/c.........}

$$(a+2d)^2 = a(a+12d)$$ ----------> $$a^2 + 4d^2 + 4ad = a^2 + 12ad$$ -----------> d^2 = 2ad -----------> d(d-2a)=0---------> Statement 1

the sum of the fourth and seventh terms of this arithmetic progression is 40 -------> (a+3d) + (a+6d) = 40 --------> 2a + 9d = 40 --------> $$d = \frac{40 - 2a}{9}$$

Putting $$d = \frac{40 - 2a}{9}$$ in the Statement 1, we get $$(\frac{40 - 2a}{9})$$$$(\frac{40 - 2a}{9} - 2a)$$ -------------------> $$(\frac{40 - 2a}{9})$$$$(\frac{40 - 20a}{9})$$ ----------> $$\frac{(40-2a)(40-20a)}{81} = 0$$ ------------------> Either (40-2a)=0 ------> a = 20 or (40-20a)=0 -----> a = 2, which is the answer
_________________
Director
Joined: 23 Jan 2013
Posts: 598
Schools: Cambridge'16
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

26 Oct 2014, 23:53
a1+3d+a1+6d=40
2a1+9d=40

a1=20-9d/2

all answer choices are integers, so 9d/2 should be integer less than 20 and 9d is even.
Only d=4 gives 18 to get 20-18=2

A
VP
Joined: 07 Dec 2014
Posts: 1067
If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

06 Aug 2016, 19:44
equation1: (a+2d)/a=(a+12d)/(a+2d)➡d=2a
equation2: 2a+9d=40
substituting,
2a+18a=40
a=2
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2649
GRE 1: Q169 V154
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

12 Aug 2016, 09:00
1
Its not that difficult as its seems

Here => a+3d+a+6d=40 => 2a+9d=40 ---> equation 1

putting d=2a in equation 1
a=2

Smash that A
_________________

MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!

Getting into HOLLYWOOD with an MBA!

The MOST AFFORDABLE MBA programs!

STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)

AVERAGE GRE Scores At The Top Business Schools!

Non-Human User
Joined: 09 Sep 2013
Posts: 7728
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

### Show Tags

12 Dec 2017, 05:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________
Re: If the first, third and thirteenth terms of an arithmetic &nbs [#permalink] 12 Dec 2017, 05:22
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.