If a₁, a₂, a₃ is any positive real numbers, then which of the following statements is not true.
(a) 3a₁, a₂, a₃≤ a₁³ + a₂³ + a₃³
(b) a₁/a₂ + a₂/a₃ + a₃/a₁≥3
(c) (a₁ + a₂ + a₃)(1/a₁ + 1/a₂ + 1/a₃)≥9
(d) (a₁ + a₂ + a₃)(1/a₁ + 1/a₂ + 1/a₃)³≤27 [JEE (Orissa)2002]
Answers
Step-by-step explanation:
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GMAT Club Forum Index Problem Solving (PS)
Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a : Problem Solving (PS)
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Bunuel
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Apr 6 at 08:18am
00:00ABCDE
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67% (03:02) correct
33% (03:06) wrong
Let , , ... , be positive integers such that . Suppose, their arithmetic mean is one less than the arithmetic mean of , , ..., . If , then the largest possible value of is
A. 20
B. 23
C. 45
D. 48
E. 50
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lacktutor
Apr 6 at 10:14am
Let , , ... , be positive integers such that <<...<. Suppose, their arithmetic mean is one less than the arithmetic mean of , , ..., . If , then the largest possible value of is
-->
In order to be the largest possible value,
-->
Answer (B).
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chetan2u
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Apr 8 at 08:21pm
Bunuel wrote:
Let , , ... , be positive integers such that . Suppose, their arithmetic mean is one less than the arithmetic mean of , , ..., . If , then the largest possible value of is
A. 20
B. 23
C. 45
D. 48
E. 50
Are You Up For the Challenge: 700 Level Questions
Sometimes understanding can make a question so simple..
Now, the crux is ..
You remove , and you get the average of remaining 51 numbers down by 1.
Thus this number has to be 51+1(itself) less than the average.
If we want to be the maximum, the average of remaining 51 should be the highest, and for that all numbers should be maximum possible and that is they should be consecutive.
So ..
Average of consecutive numbers = .
B