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In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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21 Apr 2010, 05:49
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In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5a_2=12\), what is the value of a_1? A. 4 B. 24/7 C. 2 D. 12/7 E. 6/7
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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21 Apr 2010, 06:37



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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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16 Feb 2011, 05:09
This is clearly a geometric progression. The nth term of a GP is given by a1 r^(n1) where r is the ratio between two successive terms a5 = a1 2^(51) = a1 2^4 = 16a1 a2 = a1 2^(21) = a1 2^1 = 2a1 a5  a2 = 12 16a1  2a1 = 12 14a1 = 12 a1 = 12/14 a1 = 6/7
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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17 Feb 2011, 05:49
Baten80 wrote: In the infinite sequence a1, a2, a3,...., an, each term after the first is equal to twice the previous term. If a5a2=12, what is the value of a1?
A. 4 B.24/7 C.2 D.12/7 E.6/7 Let the \(a1= x\) therefore, \(a2=2x, a3=4x, a4=8x, a5=16x.\) It is given that \(a5a2=12\), that means: \(16x2x=12\); \(14x=12\), therefore \(x=\frac{6}{7}= a1.\) Answer is E.
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In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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17 Jun 2015, 02:32
ITC wrote: In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5a_2=12\), what is the value of a_1?
A. 4 B. 24/7 C. 2 D. 12/7 E. 6/7 Just plug in, start with C: \(a_1\) to \(a_5\): 2 4 8 16 32 >>> does 324 equal 12? No! It's way too big. From C move down and repeat. E finally fits: 6/7 12/7 24/7 48/7 96/7 >>> 96/7  12/7 = 84/7 = 12
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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12 Jun 2017, 18:19
16a12a1 = 12 gives a straight forward answer as 6/7



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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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24 Feb 2018, 15:30
Bunuel wrote: In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5a_2=12\), what is the value of a_1?
A. 4 B. 24/7 C. 2 D. 12/7 E. 6/7
The formula for calculating \(n_{th}\) term would be \(a_n=2^{n1}*a_1\) . So: \(a_5=2^4*a_1\); \(a_2=2*a_1\);
Given: \(a_5a_2=2^4*a_12*a_1=12\) > \(2^4*a_12*a_1=12\) > \(a_1=\frac{12}{14}=\frac{6}{7}\).
Answer: E.
Hope it's clear. Can someone help me out here? I dont know if I am reading the question correctly but it says each term after the first term is equal to twice the previous term but the solution above shows it as two times the previous term times the first term. I am not seeing how that makes sense? Shouldnt it be \(a_5=2^4?\)



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In the infinite sequence a1, a2, a3, ..., an, each term after the firs
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24 Feb 2018, 20:47
teamryan15 wrote: Bunuel wrote: In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5a_2=12\), what is the value of a_1?
A. 4 B. 24/7 C. 2 D. 12/7 E. 6/7
The formula for calculating \(n_{th}\) term would be \(a_n=2^{n1}*a_1\) . So: \(a_5=2^4*a_1\); \(a_2=2*a_1\);
Given: \(a_5a_2=2^4*a_12*a_1=12\) > \(2^4*a_12*a_1=12\) > \(a_1=\frac{12}{14}=\frac{6}{7}\).
Answer: E.
Hope it's clear. Can someone help me out here? I dont know if I am reading the question correctly but it says each term after the first term is equal to twice the previous term but the solution above shows it as two times the previous term times the first term. I am not seeing how that makes sense? Shouldnt it be \(a_5=2^4?\) teamryan15 I think I see where you're a little off. The first term is not 1. You are focused on just the coefficient / multiplier, I think. If \(A_1\) were 1, then yes, \(A_5\) would = \(2^4\). We would have: \(A_1 = 1\) \(A_2 = 2\) \(A_3 = 4\) \(A_4 = 8\) \(A_5 = 16 = 2^4\) The first term is \(a\), not 1. Thus: \(A_1 = a_1\) \(A_2 = (2*a_1) = 2a_1 = 2^1*a_1\) \(A_3 = (2*2a_1)= 4a_1 = 2^2*a_1\) \(A_4 = (2*4a_1)= 8a_1 = 2^3* a_1\) \(A_5 = (2*8a_1) = 16a_1 = 2^4*a_1\) Hope that helps.




In the infinite sequence a1, a2, a3, ..., an, each term after the firs &nbs
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