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02 Mar 2012, 21:12
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E
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Difficulty:
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83% (01:47) correct
17%
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The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=4a_{n-1}-3\) for all integers n>1. If \(a_3\)=x, then \(a_1=\)?
A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16
A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16
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02 Mar 2012, 21:29
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The sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_n=4a_{n-1}-3\) for all integers n>1. If \(a_3\)=x, then \(a_1=\)?
A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16
Since, \(a_n=4a_{n-1}-3\) then \(a_3=4a_{2}-3\) --> \(x=4a_{2}-3\) --> \(a_2=\frac{x+3}{4}\).
Similarly, \(a_2=4a_{1}-3\) --> \(\frac{x+3}{4}=4a_{1}-3\) --> \(a_1=\frac{x+15}{16}\).
Answer: E.
Or substitute the value for \(x\), say \(x=5\), then \(a_3=5=4a_{2}-3\) --> \(a_2=2\) --> \(a_2=2=4a_{1}-3\) --> \(a_1=\frac{5}{4}\). Now, just plug \(x=5\) in the answer choices and see which one yields \(\frac{5}{4}\): only E.
Answer: E.
Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick \(x=1\) then you get three "correct" options A, C and E. Generally -1, 0, and 1 are not good choices for plug-in method.
Hope it helps.
A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16
Since, \(a_n=4a_{n-1}-3\) then \(a_3=4a_{2}-3\) --> \(x=4a_{2}-3\) --> \(a_2=\frac{x+3}{4}\).
Similarly, \(a_2=4a_{1}-3\) --> \(\frac{x+3}{4}=4a_{1}-3\) --> \(a_1=\frac{x+15}{16}\).
Answer: E.
Or substitute the value for \(x\), say \(x=5\), then \(a_3=5=4a_{2}-3\) --> \(a_2=2\) --> \(a_2=2=4a_{1}-3\) --> \(a_1=\frac{5}{4}\). Now, just plug \(x=5\) in the answer choices and see which one yields \(\frac{5}{4}\): only E.
Answer: E.
Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick \(x=1\) then you get three "correct" options A, C and E. Generally -1, 0, and 1 are not good choices for plug-in method.
Hope it helps.
02 Mar 2012, 21:34
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The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?
My Line of thought:
1. What is given to me a(3)=x, and formula.
2. I need to express a(1) from known.
3. a(n) can be expressed as a(n-1) means it can be expressed by any of the predecessor or successor.
as a(n-1) can be expressed as a(n-2)....hence a(n) can be expressed as a(n-2) and vice versa
4. a(3) = 4a(2)-3 , (n=3)>1
= 4 [4a(1)-3]-3 , (n=2)>1
=16a(1)-12-3
x = 16a(1)-15
(x+15)/16 = a(1) == Ans E
My Line of thought:
1. What is given to me a(3)=x, and formula.
2. I need to express a(1) from known.
3. a(n) can be expressed as a(n-1) means it can be expressed by any of the predecessor or successor.
as a(n-1) can be expressed as a(n-2)....hence a(n) can be expressed as a(n-2) and vice versa
4. a(3) = 4a(2)-3 , (n=3)>1
= 4 [4a(1)-3]-3 , (n=2)>1
=16a(1)-12-3
x = 16a(1)-15
(x+15)/16 = a(1) == Ans E
