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A sequence of numbers \(a_1\), \(a_2\), \(a_3\),…. is defined as follows: \(a_{2n} = a_2*a_n +1\) and \(a_{2n+1} = a_2*a_n -2\). If \(a_7 =2\) and \(0<a_1 <1\), then what is the value of \(a_{25}\)?
A. 17
B. 21
C. 82
D. 83
E. 84
M37-19
A. 17
B. 21
C. 82
D. 83
E. 84
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M37-19
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19 Nov 2021, 09:32
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Official Solution:
A sequence of numbers \(a_1\), \(a_2\), \(a_3\), … is defined as follows: \(a_{2n} = a_2*a_n +1\) and \(a_{2n+1} = a_2*a_n -2\). If \(a_7 =2\) and \(0 < a_1 < 1\), then what is the value of \(a_{25}\)?
A. \(17\)
B. \(21\)
C. \(82\)
D. \(83\)
E. \(84\)
We ned to find the value of \(a_{25}\):
\(a_{25}=a_2*a_{12} -2=a_2*(a_2*a_6 +1) -2\). So, we need the values of \(a_2\) and \(a_6\).
Notice that \(a_{2n} = a_2*a_n +1\) is the formula for even numbered terms in the sequence (\(a_2\), \(a_4\), \(a_6\), … ) and \(a_{2n+1} = a_2*a_n -2\) is the formula for odd numbered terms in the sequence (\(a_3\), \(a_5\), \(a_7\), … ).
Next, \(a_{2n} \) and \(a_{2n+1}\) are consecutive terms in the sequence, so even numbered terms are 3 more than the following odd numbered terms: \(a_{2n} -a_{2n+1}=(a_2*a_n +1) -(a_2*a_n -2)=3\). So, \(a_{2n}=a_{2n+1}+3\).
According to the above, since \(a_7 =2\), then \(a_6 =a_7+3=2+3=5\).
\(a_6 =a_2*a_3+1=5\) and since \(a_3\) must be 3 less than \(a_2\), then we'd have \(a_2*(a_2-3)+1=5\).
\((a_2)^2-3a_2-4=0\).
\((a_2-4)(a_2+1)=0\).
\(a_2=4\) or \(a_2=-1\). But from \(a_{2} = a_2*a_1 +1\) we can get that \(a_{2} =\frac{1}{1-a_1}\) and since \(0 < a_1 < 1\), then \(a_2\) must be positive, so \(a_2=4\).
Substitute \(a_2=4\) and \(a_6 =5\) into \(a_{25}=a_2*(a_2*a_6 +1) -2=4(4*5+1)-2=82\).
Answer: C
A sequence of numbers \(a_1\), \(a_2\), \(a_3\), … is defined as follows: \(a_{2n} = a_2*a_n +1\) and \(a_{2n+1} = a_2*a_n -2\). If \(a_7 =2\) and \(0 < a_1 < 1\), then what is the value of \(a_{25}\)?
A. \(17\)
B. \(21\)
C. \(82\)
D. \(83\)
E. \(84\)
We ned to find the value of \(a_{25}\):
\(a_{25}=a_2*a_{12} -2=a_2*(a_2*a_6 +1) -2\). So, we need the values of \(a_2\) and \(a_6\).
Notice that \(a_{2n} = a_2*a_n +1\) is the formula for even numbered terms in the sequence (\(a_2\), \(a_4\), \(a_6\), … ) and \(a_{2n+1} = a_2*a_n -2\) is the formula for odd numbered terms in the sequence (\(a_3\), \(a_5\), \(a_7\), … ).
Next, \(a_{2n} \) and \(a_{2n+1}\) are consecutive terms in the sequence, so even numbered terms are 3 more than the following odd numbered terms: \(a_{2n} -a_{2n+1}=(a_2*a_n +1) -(a_2*a_n -2)=3\). So, \(a_{2n}=a_{2n+1}+3\).
According to the above, since \(a_7 =2\), then \(a_6 =a_7+3=2+3=5\).
\(a_6 =a_2*a_3+1=5\) and since \(a_3\) must be 3 less than \(a_2\), then we'd have \(a_2*(a_2-3)+1=5\).
\((a_2)^2-3a_2-4=0\).
\((a_2-4)(a_2+1)=0\).
\(a_2=4\) or \(a_2=-1\). But from \(a_{2} = a_2*a_1 +1\) we can get that \(a_{2} =\frac{1}{1-a_1}\) and since \(0 < a_1 < 1\), then \(a_2\) must be positive, so \(a_2=4\).
Substitute \(a_2=4\) and \(a_6 =5\) into \(a_{25}=a_2*(a_2*a_6 +1) -2=4(4*5+1)-2=82\).
Answer: C
General Discussion
28 Aug 2021, 07:32
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Bunuel
The terms are given by
\(a_{2n} = a_2*a_n +1\)……(i)
\(a_{2n+1} = a_2*a_n -2\)….(ii)
Also, the above two terms are consecutive.
Subtract (ii) from (i) to get \(a_{2n}-a_{2n+1}=1-(-2)=3\), that is difference in consecutive integers, where even is smaller is 3.
Since \(a_7=2\), \(a_6\) will be 2+3=5
We can also find a_6 by
Quote:
Also, from the difference between consecutive terms as 3, we get \(a_2=a_3+3\)
\(a_2*a_3=4…….a_2(a_2-3)=4……(a_2)^2-3a_2-4=0……..(a_2-3)(a_2+1)=0\)
We know 0<a<1 and \(a_2=a_2*a_1+1………a_2(1-a_1)=1\), a_2 will be positive, and hence \(a_2=4.\)
Let us calculate \(a_{12}\)
\(a_{12}=a_{2*6} = a_2*a_6 +1=4*5+1=21\)
We can calculate \(a_{25}\) from it.
\(a_{2*12+1}=a_{25} = a_2*a_{12} -2=4*21-2=82\)
C
