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A geometric sequence is one in which the ratio of any term after the f
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11 Nov 2014, 09:16
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64% (01:37) correct 36% (01:50) wrong based on 178 sessions
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Tough and Tricky questions: Sequences. 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 Kudos for a correct solution.
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Re: A geometric sequence is one in which the ratio of any term after the f
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11 Nov 2014, 12:37
Bunuel wrote: Tough and Tricky questions: Sequences. 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 Kudos for a correct solution.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+c4)k... or k(3k(a+c4))=0 Clearly it does not meet criteria for GP.. Ans is D
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Re: A geometric sequence is one in which the ratio of any term after the f
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Updated on: 11 Nov 2014, 16:34
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 )
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Originally posted by clipea12 on 11 Nov 2014, 15:53.
Last edited by clipea12 on 11 Nov 2014, 16:34, edited 1 time in total.



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Re: A geometric sequence is one in which the ratio of any term after the f
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12 Nov 2014, 00:41
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
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Re: A geometric sequence is one in which the ratio of any term after the f
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12 Nov 2014, 03:58
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, \(\frac{b}{a} = \text{some constant}\), 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
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Re: A geometric sequence is one in which the ratio of any term after the f
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20 Dec 2018, 14:14
Bunuel 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 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
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.
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Re: A geometric sequence is one in which the ratio of any term after the f
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27 Feb 2019, 10:24
Bunuel wrote: Tough and Tricky questions: Sequences. 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 Kudos for a correct solution.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
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Re: A geometric sequence is one in which the ratio of any term after the f
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