Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 17 Oct 2016
Posts: 93
Location: India
Concentration: General Management, Healthcare
GMAT 1: 640 Q40 V38 GMAT 2: 680 Q48 V35
GPA: 3.05
WE: Pharmaceuticals (Health Care)

A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
29 Sep 2017, 04:05
Question Stats:
63% (01:02) correct 37% (01:18) wrong based on 126 sessions
HideShow timer Statistics
A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series? A. 1:1 B. 1:2 C. 1:4 D. 2:1 E. 4:1
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Sometimes you have to burn yourself to the ground before you can rise like a phoenix from the ashes.
Dr. Pratik



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1143
Location: India
GPA: 3.82

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
29 Sep 2017, 04:49
fitzpratik wrote: A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?
A. 1:1
B. 1:2
C. 1:4
D. 2:1
E. 4:1 Given \(a=2ar\), or \(r=\frac{1}{2}\) Hence the terms are \(a\), \(\frac{a}{2}\) & \(\frac{a}{4}\) Ratio \(= (a+\frac{a}{2})/(\frac{a}{2}+\frac{a}{4})=\frac{2}{1}\) Option D



Manager
Joined: 13 Apr 2016
Posts: 54
Location: India
GPA: 2.99
WE: Information Technology (Retail Banking)

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
08 Oct 2017, 12:29
fitzpratik wrote: A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?
A. 1:1
B. 1:2
C. 1:4
D. 2:1
E. 4:1 The ans should be 2:1. Though I read the question incorrect as the first term being half of the second term.



Intern
Joined: 28 Aug 2016
Posts: 16

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
14 Nov 2017, 16:18
Wouldn't ar^2 become a^2/4?



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1143
Location: India
GPA: 3.82

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
14 Nov 2017, 19:44
BigUD94 wrote: Wouldn't ar^2 become a^2/4? Hi BigUD94Here it is r^2 and not a^2, hence it will be a/4 when you substitute the value of r. Posted from my mobile device



Intern
Joined: 28 Aug 2016
Posts: 16

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
15 Nov 2017, 09:06
niks18 Oh alright sir I thought it was (ar)^2 by mistake, I got it now. Thanks



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1143
Location: India
GPA: 3.82

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
15 Nov 2017, 09:24
BigUD94 wrote: niks18 Oh alright sir I thought it was (ar)^2 by mistake, I got it now. Thanks Hi BigUD94you suddenly made me 50 year old guy by writing that "sir"



Math Expert
Joined: 02 Aug 2009
Posts: 5938

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
15 Nov 2017, 21:21
fitzpratik wrote: A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?
A. 1:1
B. 1:2
C. 1:4
D. 2:1
E. 4:1 Such Questions can be done in two ways.. Algebraic as shown above \(2ar=a......r=\frac{1}{2}\)... So terms are a, a/2 and a/4.. Now (a+a/2)/(a/2+a/4)=(3a/2)/(3a/4)=4/2=2/1... 2:1 Other slightly easier is substitution, rather smart substitution looking at the question Let a=4.. So ar will be half, 4/2=2 And ar^2 will become 2/2=1 Number 4,2,1... We are looking for \(\frac{4+2}{2+1}=\frac{6}{3}=\) So 2:1
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
GMAT online Tutor



Intern
Joined: 28 Aug 2016
Posts: 16

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
17 Nov 2017, 11:50
niks18 Har koi gyaani hai idhar, mu se sir nikal hi jaata hai out of respect :D



SC Moderator
Joined: 22 May 2016
Posts: 1762

A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
17 Nov 2017, 17:35
chetan2u wrote: fitzpratik wrote: A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?
A. 1:1
B. 1:2
C. 1:4
D. 2:1
E. 4:1 Such Questions can be done in two ways.. Algebraic as shown above \(2ar=a......r=\frac{1}{2}\)... So terms are a, a/2 and a/4.. Now (a+a/2)/(a/2+a/4)=(3a/2)/(3a/4)=4/2=2/1... 2:1 Other slightly easier is substitution, rather smart substitution looking at the question Let a=4.. So ar will be half, 4/2=2 And ar^2 will become 2/2=1 Number 4,2,1... We are looking for \(\frac{4+2}{2+1}=\frac{6}{3}=\) So 2:1 chetan2u , I like this second method. I'm a little confused by this part: Quote: Let a=4.. So ar will be half, 4/2=2 And ar^2 will become 2/2=1 Why is third term derived from 2/2? I thought it would be: \(a = 4\) \(a_2 = 4 * \frac{1}{2} = 2\) \(a_3 = 4 * \frac{1}{4} = 1\) I'm missing something about what happened to ratio, r, here . . .
_________________
In the depths of winter, I finally learned that within me there lay an invincible summer.  Albert Camus, "Return to Tipasa"



Math Expert
Joined: 02 Aug 2009
Posts: 5938

Re: A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
17 Nov 2017, 19:37
genxer123 wrote: chetan2u wrote: fitzpratik wrote: A series has three numbers, a, ar and ar^2. In the series, the first term is twice the second term. What is the ratio of sum of the first two terms to the sum of the last two terms in series?
A. 1:1
B. 1:2
C. 1:4
D. 2:1
E. 4:1 Such Questions can be done in two ways.. Algebraic as shown above \(2ar=a......r=\frac{1}{2}\)... So terms are a, a/2 and a/4.. Now (a+a/2)/(a/2+a/4)=(3a/2)/(3a/4)=4/2=2/1... 2:1 Other slightly easier is substitution, rather smart substitution looking at the question Let a=4.. So ar will be half, 4/2=2 And ar^2 will become 2/2=1 Number 4,2,1... We are looking for \(\frac{4+2}{2+1}=\frac{6}{3}=\) So 2:1 chetan2u , I like this second method. I'm a little confused by this part: Quote: Let a=4.. So ar will be half, 4/2=2 And ar^2 will become 2/2=1 Why is third term derived from 2/2? I thought it would be: \(a = 4\) \(a_2 = 4 * \frac{1}{2} = 2\) \(a_3 = 4 * \frac{1}{4} = 1\) I'm missing something about what happened to ratio, r, here . . . Hi.. No you are correct, both mean the same thing... you have taken \(\frac{4}{4}=4*\frac{1}{4}\) as \(a*r^2\), that is first term * r^2 whereas in my calculations ... \(\frac{2}{2}=2*\frac{1}{2}\) as \(ar*r\), that is 2nd term * r
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
GMAT online Tutor



SC Moderator
Joined: 22 May 2016
Posts: 1762

A series has three numbers, a, ar and ar^2. In the series, the first.. [#permalink]
Show Tags
17 Nov 2017, 20:48
chetan2u wrote: Hi..
No you are correct, both mean the same thing... you have taken \(\frac{4}{4}=4*\frac{1}{4}\) as \(a*r^2\), that is first term * r^2 whereas in my calculations ... \(\frac{2}{2}=2*\frac{1}{2}\) as \(ar*r\), that is 2nd term * r chetan2u , thanks! My brain shortcircuited. I appreciate the help.
_________________
In the depths of winter, I finally learned that within me there lay an invincible summer.  Albert Camus, "Return to Tipasa"




A series has three numbers, a, ar and ar^2. In the series, the first..
[#permalink]
17 Nov 2017, 20:48






