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27 Jun 2018, 02:02
Kudos
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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:
35%
(medium)
Question Stats:
75% (02:23) correct
25%
(02:30)
wrong
based on 153
sessions
History
Date
Time
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Not Attempted Yet
[GMAT math practice question]
If \(A_{n+1}=120 + (\frac{1}{4})A_n\) for each positive integer \(n\), and \(A_1=150\), which of these ranges of values includes \(A_{12}\)?
A. 110-120
B. 120-130
C. 130-140
D. 140-150
E. 150-160
If \(A_{n+1}=120 + (\frac{1}{4})A_n\) for each positive integer \(n\), and \(A_1=150\), which of these ranges of values includes \(A_{12}\)?
A. 110-120
B. 120-130
C. 130-140
D. 140-150
E. 150-160
27 Jun 2018, 03:19
Kudos
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MathRevolution
\(A_2 = 120 + \frac{1}{4}A_1 = 120 + \frac{1}{4}*150 = 120 + 37.5 = 157.5\)
\(A_3 = 120 + \frac{1}{4}A_2 = 120 + \frac{1}{4}*157.5 ≈ 120 + 39 = 159\)
\(A_4 = 120 + \frac{1}{4}A_3 = 120 + \frac{1}{4}*159\)...
The terms are increasing in value.
Thus, A₁₂ > 159.
27 Jun 2018, 03:24
Kudos
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MathRevolution
We are given An+1=\(120 + (\frac{1}{4})\)An
A2 = \(120 + (\frac{1}{4})*150 = \frac{630}{4}\) (As A1=150)
A3 = \(120 + (\frac{1}{4}\))A2 = \(120 + (\frac{1}{4} * \frac{630}{4}) = 120 + 37.5 = 157.5\) (As A2=\(\frac{630}{4}\))
We can repeat this process until A12 and the value will only increase from this value.
Therefore, the range of the values that can be a possible value for A12 is Option E(150 - 160)
