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31 Mar 2021, 08:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
97% (01:21) correct
3%
(02:01)
wrong
based on 89
sessions
History
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If \(A_{n} = 2A_{n−1} + 3\) for all \(n ≥ 1\), and \(A_{4} = 45\), what is \(A_{1}\)?
(A) 45
(B) 21
(C) 9
(B) 6
(E) 3
(A) 45
(B) 21
(C) 9
(B) 6
(E) 3
31 Mar 2021, 08:51
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2A3+3=45 A3=21
2A2+3=21 A2=9
2A1+3=9 A1=3
Option E is the correct choice
2A2+3=21 A2=9
2A1+3=9 A1=3
Option E is the correct choice
31 Dec 2024, 01:03
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Given that \(A_{n} = 2A_{n−1} + 3\) for all \(n ≥ 1\) and \(A_{4} = 45\) and we need to find the value of \(A_{1}\))
Let \(A_{n} = 2A_{n−1} + 3\) be denoted by (1)
So, we need to use the above equation and try to calculate \(A_{3}\) from \(A_{4}\) and then \(A_{2}\) from \(A_{3}\) and eventually \(A_{1}\) from \(A_{2}\).
Two ways of doing this
Method 1: Find value at each stage and then solve for the next stage
So, lets start with the labor work!
Putting n = 4 in (1) we get, \(A_{4} = 2A_{4−1} + 3\)
=> \(45 = 2A_{3} + 3\)
=> \(2A_{3} = 42\) => \(A_{3} = 21\)
Putting n = 3 in (1)we get, \(A_{3} = 2A_{3−1} + 3\)
=> \(21 = 2A_{2} + 3\)
=> \(2A_{2} = 18\) => \(A_{2} = 9\)
Putting n = 2 in (1) we get, \(A_{2} = 2A_{2−1} + 3\)
=> \(9 = 2A_{1} + 3\)
=> \(2A_{1} = 6\) => \(A_{1} = 3\)
So, Answer will be E.
Method 2: Simplify the entire expression first and then solve for the value
Putting n = 4 in (1) we get, \(A_{4} = 2A_{4−1} + 3\) => \(A_{4} = 2A_{3} + 3\), Now let's find \(A_{3}\)
Putting n=3 in (1) we get, \(A_{3} = 2A_{3−1} + 3\) => \(A_{3} = 2A_{2} + 3\)
Similarly, \(A_{2} = 2A_{1} + 3\)
=> \(A_{4} = 2A_{3} + 3\) will give us \(A_{4} = 2(2A_{2} + 3) + 3\) => \(A_{4} = 4A_{2} + 9\)
And this will give us \(A_{4} = 4(2A_{1} + 3) + 9\) => \(A_{4} = 8A_{1} + 21\)
=> \(45 = 8A_{1} + 21\)
=> \(8A_{1} = 24\)
=> \(A_{1} = 3\)
So, Answer will be E.
Hope it helps!
Watch the following video to learn How to Sequence problems
Let \(A_{n} = 2A_{n−1} + 3\) be denoted by (1)
So, we need to use the above equation and try to calculate \(A_{3}\) from \(A_{4}\) and then \(A_{2}\) from \(A_{3}\) and eventually \(A_{1}\) from \(A_{2}\).
Two ways of doing this
Method 1: Find value at each stage and then solve for the next stage
So, lets start with the labor work!
Putting n = 4 in (1) we get, \(A_{4} = 2A_{4−1} + 3\)
=> \(45 = 2A_{3} + 3\)
=> \(2A_{3} = 42\) => \(A_{3} = 21\)
Putting n = 3 in (1)we get, \(A_{3} = 2A_{3−1} + 3\)
=> \(21 = 2A_{2} + 3\)
=> \(2A_{2} = 18\) => \(A_{2} = 9\)
Putting n = 2 in (1) we get, \(A_{2} = 2A_{2−1} + 3\)
=> \(9 = 2A_{1} + 3\)
=> \(2A_{1} = 6\) => \(A_{1} = 3\)
So, Answer will be E.
Method 2: Simplify the entire expression first and then solve for the value
Putting n = 4 in (1) we get, \(A_{4} = 2A_{4−1} + 3\) => \(A_{4} = 2A_{3} + 3\), Now let's find \(A_{3}\)
Putting n=3 in (1) we get, \(A_{3} = 2A_{3−1} + 3\) => \(A_{3} = 2A_{2} + 3\)
Similarly, \(A_{2} = 2A_{1} + 3\)
=> \(A_{4} = 2A_{3} + 3\) will give us \(A_{4} = 2(2A_{2} + 3) + 3\) => \(A_{4} = 4A_{2} + 9\)
And this will give us \(A_{4} = 4(2A_{1} + 3) + 9\) => \(A_{4} = 8A_{1} + 21\)
=> \(45 = 8A_{1} + 21\)
=> \(8A_{1} = 24\)
=> \(A_{1} = 3\)
So, Answer will be E.
Hope it helps!
Watch the following video to learn How to Sequence problems
