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01 Jul 2022, 01:30
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If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is
A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)
Are You Up For the Challenge: 700 Level Questions
A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)
If positive real numbers a,b,c are in a.p such that abc=4 Then the minimum value of b is
Are You Up For the Challenge: 700 Level Questions
01 Jul 2022, 05:51
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We are told, \(abc = 4\). Looking at the answer choices all are exponents of 2, so let \(abc = 2^2\)
As we are dealing with an arithmetic sequence we know that each term in the sequence will have the same base and the difference between them each will be a constant. Let
Let \(a = 2^n\), \(b = 2^{n+x}\) & \(c=2^{n+2x}\)
Therefore \((2^n)*(2^{n+x})*2^{n+2x} = 2^2\)
\(2^{n+n+x+n+2x} = 2^2\)
Drop the bases
\(3n + 3x = 2\)
What we know is that \(b = 2^{n+x}\) so to get \(n+x\) we divide the above equation through by 3 to get:
\(n+x = \frac{2}{3}\)
So \(b = 2^{\frac{2}{3}}\)
Answer B
As we are dealing with an arithmetic sequence we know that each term in the sequence will have the same base and the difference between them each will be a constant. Let
Let \(a = 2^n\), \(b = 2^{n+x}\) & \(c=2^{n+2x}\)
Therefore \((2^n)*(2^{n+x})*2^{n+2x} = 2^2\)
\(2^{n+n+x+n+2x} = 2^2\)
Drop the bases
\(3n + 3x = 2\)
What we know is that \(b = 2^{n+x}\) so to get \(n+x\) we divide the above equation through by 3 to get:
\(n+x = \frac{2}{3}\)
So \(b = 2^{\frac{2}{3}}\)
Answer B
24 Jul 2022, 06:47
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Bunuel
If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is
A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)
A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)
kapil1995, I believe the solution above is not correct as \(2^n,2^{n+x},2^{n+2x}\) is \(2^n,2^x(2^n),2^x(2^x2^n)\), that is successive term is getting multiplied by \(2^x\), making the sequence Geometric progression.
Logical:
a*b*c=4 and looking at the options
Since the terms are equidistant, the way to minimise b and c as also maximise a is by taking all three same => a=b=c
Thus, \(a*b*c=b^3=4=2^2………b=2^{\frac{2}{3}}\)
Algebraic method
Now, you could also verify that we have to look for \(b^3=2^2\)
Let the product be \((b-x)*b*(b+x)=b(b^2-x^2)\)
Now \(b^2-x^2<b^2\) unless x is 0.
If x can be 0 too, then \(b^2-x^2\leq b^2\)
Multiply both sides by b without changing the sign as b is positive
\(b(b^2-x^2)\leq b(b^2)\), but \(b(b^2-x^2)=4\)
Thus, \(4\leq b^3……2^2\leq b^3\)
Taking cube root,
\(2^{\frac{2}{3}}\leq b\)
Thus, b is at least \(2^{\frac{2}{3}}\)
