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

 It is currently 17 Oct 2019, 18:34 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If the first, third and thirteenth terms of an arithmetic

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
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

2
12 00:00

Difficulty:   55% (hard)

Question Stats: 70% (02:53) correct 30% (02:44) wrong based on 156 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 V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

13
4
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!
The answer is (A)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
General Discussion
Director  Joined: 22 Mar 2011
Posts: 588
WE: Science (Education)
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

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

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 $$a_1=2.$$

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

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: 39
Location: Austria
GMAT 1: 770 Q51 V42 Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

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: 2
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

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: 127
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

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 V
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
MBA Section Director V
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 7111
City: Pune
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

4
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
_________________
2020 MBA Applicants: Introduce Yourself Here!

MBA Video Series - Video answers to specific components and questions about MBA applications.

2020 MBA Deadlines, Essay Questions and Analysis of all top MBA programs
Director  G
Joined: 23 Jan 2013
Posts: 525
Schools: Cambridge'16
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

1
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  P
Joined: 07 Dec 2014
Posts: 1222
If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

equation1: (a+2d)/a=(a+12d)/(a+2d)➡d=2a
equation2: 2a+9d=40
substituting,
2a+18a=40
a=2
Current Student D
Joined: 12 Aug 2015
Posts: 2568
Schools: Boston U '20 (M)
GRE 1: Q169 V154 Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

1
Its not that difficult as its seems

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

hence 8ad=4d^2 => d=2a
putting d=2a in equation 1
a=2

Smash that A
_________________
Intern  B
Joined: 16 Oct 2016
Posts: 5
Re: If the first, third and thirteenth terms of an arithmetic  [#permalink]

Show Tags

Given 1st,3rd and 13th term in GP.
so $$b^{2}$$=ac (GP Mean rule)
$$t_3$$$$^{2}$$=$$t_1$$*$$t_{13}$$
$$(a+2d)^{2}$$=a(a+12d)
$$a^{2}$$+4ad+4$$d^{2}$$=$$a^{2}$$+12ad
4$$d^{2}$$=8ad
4d(d-2a)=0 (d is not equal to zero)
d=2a------------- (1)

Given $$t_4$$+$$t_7$$=40
a+3d+a+6d=40
2a+9(2a)=40 from (1)
20a=40
First term a =2 Re: If the first, third and thirteenth terms of an arithmetic   [#permalink] 08 Feb 2019, 03:38
Display posts from previous: Sort by

If the first, third and thirteenth terms of an arithmetic

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

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