Problem Solving

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

The OA is D.

Source: Manhattan Prep

swerve wrote: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 nonzero 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
Source: Manhattan Prep

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!

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

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

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


The correct answer is therefore (D).


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

Regards,
Fabio.