The second, the first and the third term of an AP whose comm

Question

The second, the first and the third term of an AP whose common difference is non zero but lesser than 200, form a GP in that order. What is the common ration of that GP?

A. 1
B. -1
C. 2
D. -2
E. |1|

KarishmaB · Accepted Answer

Detailed algebraic explanation:

Let the 3 terms of the AP be (a-d), a and (a+d)
Terms of the GP: a, (a-d), (a+d) in that order.
In a GP, terms next to each other have the same ratio.
So,  \frac{(a-d)}{a} = \frac{(a+d)}{(a-d)}

(a-d)^2 = a(a+d)

d^2 - 2ad = ad

d^2 - 3ad = 0

d(d - 3a) = 0

We know that d is not 0 from the question. So d = 3a

Common ratio  = \frac{(a-d)}{a} = \frac{(a - 3a)}{a} = -2

avenkatesh007 · Answer

its D

think A.P to be a-d,a,a+d.
G.P will be a,a-d,a+d

which means (a-d)^2 = a^2 + ad
that means d=3
and G.P is -1,2,-4 and A.P is 2,-1,-4. so G.P common ratio is -2 .....

Anasthaesium · Answer

Can you please elaborate on how you solved the quad with two unknowns variables.

avenkatesh007 · Answer

(a-d)^2 = a^2 +ad
a^2-2ad+d^2=a^2+ad
d^2=3ad
d=3a
here I assumed 'a' to be 1 coz in G.P ratio will be a+d:a kind of.so it wont matter what 'a' is.......else u can substitute direct values interms of 'a'.

siddharthmuzumdar · Answer

I am getting the ratio as -1/2. Can someone please explain how it is -2?

anuu · Answer

Hi,

Can you please explain how did u get:

(a-d)^2 = a^2+ad

Thanks,
Anu

subhajeet · Answer

for GP we use b^2=ac

so using that (a-d)^2 = a(a+d)

by solving this we get d=3a

but the common difference is comming to be -1/2
can anyone please comment on this.

siddharthmuzumdar · Answer

subhajeet, I have cited the same problem above. Even I am getting the ratio as -1/2. Wonder if we are missing something vital here.

KarishmaB · Answer

You probably got d = 3a but after that, substituted d in a/(a-d) as one would naturally since (a-d) is smaller than a. But, the terms in the GP are a, (a-d), (a+d) in that order. So the common ratio is (a-d)/a or (a+d)/(a-d)

subhajeet · Answer

Karishma: U got me right. I was indeed making the same mistake as you have mentioned here. Thanks for the reply.

siddharthmuzumdar · Answer

Grrr....I am just cursing myself for such silly mistakes.

Thanks a ton for pointing it out.

taleesh · Answer

Hi Karishma,

I am bit confuse with this AP and GP, is it Arithmetic progression and geometric progression. And how do we decide this sequence of a,a-d, a+d.

KarishmaB · Answer

Yes, AP is Arithmetic Progression, GP is Geometric Progression.

The second first and third terms of an AP form a GP when put in that order.

How do we express the terms of AP? Three terms can be expressed as 
a-d, a, a+d (with d as the common difference)

When you put them in this order: second, first and third
a, a-d, a+d - this is a GP

A GP has common ratio so (a-d)/a = (a+d)/(a-d) = Common Ratio

OreoShake · Answer

this does not look like a GMAT question. Please let me know if im wrong.

KarishmaB · Answer

The concept could easily be tested this way in GMAT though the wording of the question would be much more explicit.

GmatPoint · Answer

﻿Considering the first second and the third term of the A.P to be:  a,\ a+d,\ a+2\cdot d 
﻿Given that the second, first, and the third term in the given order form a geometric progression.
﻿Hence we can rewrite a = k*r, a+d = k, a+d =  kr^2 
﻿In the form of a geometric progression.
﻿Since the square of the second term is equal to the product of the first and the third term in a G.P.
﻿We have : ﻿ a^2=\ \left(a+d\right)\left(a+2d\right) .
﻿Solving this we have:  a^2=\ a^2+3ad+2d^2 .
﻿Hence d =  -\frac{3a}{2} 
﻿Substituting the value of d in the three terms we have :
﻿k =  -\frac{a}{2} 
﻿kr = a
 ﻿kr^2  = -2a
﻿Hence the value of r = -2 which is the common ratio.
﻿

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