A geometric sequence is one in which the ratio of any term after the f

Question

A geometric sequence is one in which the ratio of any term after the first to the preceding term is a constant. If the letters  a ,  b ,  c ,  d  represent a geometric sequence in normal alphabetical order, which of the following must also represent a geometric sequence for all values of  k ?
 
I.  dk ,  ck ,  bk ,  ak 
 
II.  a + k ,  b + 2k ,  c + 3k ,  d + 4k 
 
III.  ak^4 ,  bk^3 ,  ck^2 ,  dk

A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II, and III

Bunuel · Accepted Answer

Official Solution:

A geometric sequence is one in which the ratio of any term after the first to the preceding term is a constant. If the letters  a ,  b ,  c ,  d  represent a geometric sequence in normal alphabetical order, which of the following must also represent a geometric sequence for all values of  k ? 
 
I.  dk ,  ck ,  bk ,  ak 
 
II.  a + k ,  b + 2k ,  c + 3k ,  d + 4k 
 
III.  ak^4 ,  bk^3 ,  ck^2 ,  dk

A. I only
B. I and II only
C. II and III only
D. I and III only
E. I, II, and III

Our first task is to ensure that we understand the definition of a geometric sequence. Let's use the sequence given to us:  a ,  b ,  c ,  d . We are told that the ratio of any term after the first to the preceding term is a constant. In other words,  \frac{b}{a} = 	ext{some constant} , which is the same constant for the other ratios ( \frac{c}{b}  and  \frac{d}{c} ). Let's name that constant  r . Thus, we have the following:
  \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = r 
 
By a series of substitutions, we can rewrite the sequence in terms of just  a  and  r :
 
 a ,  b ,  c ,  d  is the same as  a ,  ar ,  ar^2 ,  ar^3 
 
Rewriting the sequence this way highlights the role of the constant ratio  r . That is, to move forward in the sequence one step, we just multiply by a constant factor  r . Rewriting also lets us substitute into the alternative sequences and watch what happens.
 
I.  dk ,  ck ,  bk ,  ak 
 
This sequence is the same as  ar^3k ,  ar^2k ,  ark ,  ak . To move forward in the sequence, we divide by  r . This is the same thing as multiplying by  \frac{1}{r} . Since this factor is constant throughout the sequence, the sequence is geometric.
 
II.  a + k ,  b + 2k ,  c + 3k ,  d + 4k 
 
This sequence is the same as  a + k ,  ar + 2k ,  ar^2 + 3k ,  ar^3 + 4k . To move forward in this sequence, we cannot simply multiply by a constant expression. The presence of the plus sign means that we will not have a constant ratio between successive terms, and this sequence is not geometric.
 
III.  ak^4 ,  bk^3 ,  ck^2 ,  dk 
 
This sequence is the same as  ak^4 ,  ark^3 ,  ar^2k^2 ,  ar^3k . To move forward in the sequence, we multiply by  r  and divide by  k . In other words, we multiply by  \frac{r}{k} , which is a constant factor. This sequence is therefore geometric.
 
Only sequences I and III are geometric.

Answer: D

WoundedTiger · Answer

Note that if a,b,c and d are in GP then... b^2=ac and c^2=bd

Option 1:  dk ,  ck ,  bk ,  ak  so we have c^2k^2=dk*bk...Sufficient 1 is GP...Option C ruled out
Option 3 is better bet so let's do that before: b^2*k^6=ak^4*ck^2 or b^2*k^6=ack^6...St 2 is also GP.... Option A and B are out

Option 2:  a + k ,  b + 2k ,  c + 3k ,  d + 4k

(b+2k)^2=(a+k)*(c+k) or b^2+4k^2+4k=ac+ak+ck+k^2 ---->3k^2 =(a+c-4)k... or k(3k-(a+c-4))=0 Clearly it does not meet criteria for GP..

Ans is D

clipea12 · Answer

Somehow I feel that i got tricked
First, a geometric sequence has to be on this form ax ,bx, cx... ( x is the common factor(ratio))
so 1) is definitely correct
3) dividing the sequence will make it easy to understand ak^4, bk^3 is a geometric sequence with K as a common factor. The same would be for the rest ck^2, dk^1
So 3) is correct
2) is a straight arithmetic sequence
hence D is the answer (unless i got tricked since the title is tricky questions

)

PareshGmat · Answer

a, b, c, d are in GA, then

\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = Constant

b^2 = ac

c^2 = bd

Sequence II fails. I & III are sure winner

Answer = D

fskilnik · Answer

Important: the fact that a,b,c,d is a given GP in that order guarantees, implicitly, that a,b, c and d are not zero. Think about that!

\left( {\rm{I}} \right)\,\,\,\,{{ck} \over {dk}} = {{bk} \over {ck}} = {{ak} \over {bk}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{c \over d} = {b \over c} = {a \over b}\,\,\,\,\,\,\left( { \Leftrightarrow \,\,\,\,\,\,{d \over c} = {c \over b} = {b \over a}} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,a,b,c,d\,\,\,{\rm{GP}}\,\,\,:\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{refute}}\,\,\,\,\left( {\rm{C}} \right)

\left( {{\rm{II}}} \right)\,\,\,{\rm{Take}}\,\,{\rm{GP}}\,\,\left( {a,b,c,d} \right) = \left( {1,2,4,8} \right)\,\,{\rm{and}}\,\,k = 1\,\,:\,\,\,\,\,\,\left( {2,4,7,12} \right)\,\,\,{\rm{not}}\,\,{\rm{GP}}\,\,\,\,{\rm{:}}\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refute}}\,\,{\rm{also}}\,\,\left( {\rm{B}} \right),\left( {\rm{E}} \right)

\left( {{\rm{III}}} \right)\,\,{{b{k^3}} \over {a{k^4}}} = {{c{k^2}} \over {b{k^3}}} = {{dk} \over {c{k^2}}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{b \over {ak}} = {c \over {bk}} = {d \over {ck}}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \matrix{
 \,{b \over {ak}} = {c \over {bk}}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{b \over a} = {c \over b}\,\,\,\,\,\, \Leftrightarrow \,\,\,{b \over c} = {a \over b}\,\,\,\,\,\, \hfill \cr 
 \,{c \over {bk}} = {d \over {ck}}\,\,\,\, \Leftrightarrow \,\,\,\,\,{c \over b} = {d \over c}\,\,\,\,\,\, \Leftrightarrow \,\,\,{c \over d} = {b \over c}\, \hfill \cr} \right.\,\,\, \Leftrightarrow \,\,\,\,\,a,b,c,d\,\,\,{\rm{GP}}\,\,\,:\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,

The correct answer is therefore (D).

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

ScottTargetTestPrep · Answer

Recall that in a geometric sequence, the ratio of any term after the first to the preceding term is called the common ratio. We can let a, b, c and d be 1, 2, 4, and 8, respectively, (notice that the common ratio is 2) and k be 3. Now let’s analyze each Roman numeral.

I. dk, ck, bk, ak

dk = 8(3) = 24, ck = 4(3) = 12, bk = 2(3) = 6 and ak = 1(3) = 3

The sequence is 24, 12, 6, 3. This is a geometric sequence with a common ratio of 1/2.

II. a+k, b+2k, c+3k, d+4k

a+k = 1 + 3 = 4, b+2k = 2 + 2(3) = 8, c+3k = 4 + 3(3) = 13 and d+4k = 8 + 4(3) = 20

The sequence is 4, 8, 13, 28. However, this is NOT a geometric sequence because we don’t have a nonzero constant as the common ratio.

III. ak^4, bk^3, ck^2, dk

ak^4 = 1(3)^4 = 81, bk^3 = 2(3)^3 = 54, ck^2 = 4(3)^2 = 36, dk = 8(3) = 24

The sequence is 81, 54, 36, 24. This is a geometric sequence with a common ratio of 2/3.

Answer: D

ScottTargetTestPrep · Answer

We can let a, b, c, d, and k be 1, 2, 4, 8, and 2, respectively (notice that 1, 2, 4, 8 form a geometric sequence with common ratio = 2). Now, let’s analyze each Roman numeral with the numbers we have for a, b, c, d, and k.

I. dk, ck, bk, ak = 16, 8, 4, 2

This is a geometric sequence with common ratio = 1/2.

II. a + k, b + 2k, c + 3k, d + 4k = 3, 6, 10, 16

This is not a geometric sequence since 6/3 ≠ 10/6 ≠ 16/10.

III. ak^4, bk^3, ck^2, dk = 16, 16, 16, 16

This is a geometric sequence with common ratio = 1.

Answer: D

pierjoejoe · Answer

straight geometric sequence:

a 
 b 
 c 
 d 
 
  a  
 ar  
 ar^2  
 ar^3

the third sequence, we are just multiplying each term in the straight geometric sequence by a certain k to a power
   a 
 b 
 c 
 d 
 
  ak^4  
 ark^3  
 ar^2k^2  
 ar^3k

meaning that the ratios between terms is always constant and equal to r/k

i misinterpreted, because i initially thought that

a 
 b 
 c 
 d 
 
  a  (k^4)   
  a (k^4)  (r)  (k^3)   
  a (k^7) (r) (r) (k^2)   
  a (k^9)(r^2) (r) (k)   
 
   a (k^4)   
  a (k^7) (r)   
  a (k^9)(r^2)   
 a (k^10)(r^3)

why isn't it like this?  Bunuel

bumpbot · Answer

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A geometric sequence is one in which the ratio of any term after the f

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a	b	c	d
a (k^4)	a (k^4) (r) (k^3)	a (k^7) (r)(r)(k^2)	a (k^9)(r^2)(r)(k)
a (k^4)	a (k^7) (r)	a (k^9)(r^2)	a (k^10)(r^3)

GMAT Problem Solving (PS) Questions

A geometric sequence is one in which the ratio of any term after the f

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