Question

Let \[G\] be the geometric mean of two positive numbers \[a\] and \[b\] and \[M\] be the arithmetic mean of \[\dfrac{1}{a}\] and \[\dfrac{1}{b}\]. If \[\dfrac{1}{M}:G\] is \[4:5\], then \[a:b\] can be
A. \[1:4\]
B. \[1:2\]
C. \[2:3\]
D. \[3:4\]

Answer 1

Hint: Here we use the concept of Arithmetic mean of two numbers and geometric mean of two numbers and write \[M\] and \[G\] in terms of \[a\] and \[b\]. Using the ratio given we write the ratio in fraction form and substitute the relation obtained earlier. Solve for the values of \[a\] and \[b\] and calculate their ratio.
* Arithmetic mean of two numbers \[p\] and \[q\] is given by \[\dfrac{{p + q}}{2}\].
* Geometric mean of two numbers \[p\] and \[q\] is given by \[\sqrt {pq} \]
* Here we use the concept of ratio which gives us a relation between two quantities i.e. If we say \[m:n = 1:2\], we mean for every \[1m\] there is \[2n\]
* Ratio \[m:n = 1:2\] can be written as \[\dfrac{m}{n} = \dfrac{1}{2}\].

Complete step-by-step answer:
We are given that the geometric mean of two numbers \[a\]and \[b\] is \[G\].
We know geometric mean of \[p\] and \[q\] is \[\sqrt {pq} \], Substitute \[p = a\] and \[q = b\]
\[ \Rightarrow G = \sqrt {ab} \] … (1)
Now we are given that the arithmetic mean of two numbers \[\dfrac{1}{a}\] and \[\dfrac{1}{b}\] is \[M\].
We are given that arithmetic mean of \[p\] and \[q\] is \[\dfrac{{p + q}}{2}\], Substitute \[p = \dfrac{1}{a}\] and \[q = \dfrac{1}{b}\]
\[ \Rightarrow M = \dfrac{{\dfrac{1}{a} + \dfrac{1}{b}}}{2}\]
Taking LCM in the numerator we get
\[ \Rightarrow M = \dfrac{{\dfrac{{a + b}}{{ab}}}}{2}\]
\[ \Rightarrow M = \dfrac{{a + b}}{{2ab}}\]
Taking reciprocal on both sides we get
\[ \Rightarrow \dfrac{1}{M} = \dfrac{{2ab}}{{a + b}}\] … (2)
Now we know the ration of \[\dfrac{1}{M}:G\] is \[4:5\]
\[ \Rightarrow \dfrac{{\dfrac{1}{M}}}{G} = \dfrac{4}{5}\]
Substitute the values of \[M\]and \[G\] from equation (1) and (2)
\[ \Rightarrow \dfrac{{\dfrac{{2ab}}{{a + b}}}}{{\sqrt {ab} }} = \dfrac{4}{5}\]
We can simplify and write the above fraction in numerator as
\[ \Rightarrow \dfrac{{2ab}}{{\sqrt {ab} \times (a + b)}} = \dfrac{4}{5}\]
Now we know that \[ab = \sqrt {ab} \times \sqrt {ab} \].
\[ \Rightarrow \dfrac{{2(\sqrt {ab} \times \sqrt {ab} )}}{{\sqrt {ab} \times (a + b)}} = \dfrac{4}{5}\]
Cancel out the same factors from numerator and denominator
\[ \Rightarrow \dfrac{{2\sqrt {ab} }}{{(a + b)}} = \dfrac{4}{5}\]
Cancel 2 from numerator of both sides of the equation
\[ \Rightarrow \dfrac{{\sqrt {ab} }}{{(a + b)}} = \dfrac{2}{5}\]
Now cross multiply both sides of the equation
\[ \Rightarrow 5\sqrt {ab} = 2(a + b)\]
Squaring both sides of the equation we get
\[
   \Rightarrow {\left( {5\sqrt {ab} } \right)^2} = {\left( {2(a + b)} \right)^2} \\
   \Rightarrow 25{(\sqrt {ab} )^2} = 4{(a + b)^2} \\
 \]
Cancel the square root by square power in LHS and use the formula \[{(x + y)^2} = {x^2} + {y^2} + 2xy\] in RHS of the equation.
\[
   \Rightarrow 25ab = 4({a^2} + {b^2} + 2ab) \\
   \Rightarrow 25ab = 4{a^2} + 4{b^2} + 8ab \\
 \]
Shift all the values to one side of the equation
\[
   \Rightarrow 4{a^2} + 4{b^2} + 8ab - 25ab = 0 \\
   \Rightarrow 4{a^2} + 4{b^2} - 17ab = 0 \\
 \]
Now we factorize the equation.
We can write \[ - 17ab = - 16ab - ab\]
\[
   \Rightarrow 4{a^2} - 17ab + 4{b^2} = 0 \\
   \Rightarrow 4{a^2} - 16ab - ab + 4{b^2} = 0 \\
 \]
Now we take 4a common from the first two terms and –b from last two terms.
\[
   \Rightarrow 4a(a - 4b) - b(a - 4b) = 0 \\
   \Rightarrow (4a - b)(a - 4b) = 0 \\
 \]
Equate each factor to zero.
Firstly,
\[ \Rightarrow 4a - b = 0\]
Shift b to other side of the equation
\[ \Rightarrow 4a = b\]
Divide both sides by 4b
\[ \Rightarrow \dfrac{{4a}}{{4b}} = \dfrac{b}{{4b}}\]
Cancel out same terms from both numerator and denominator
\[ \Rightarrow \dfrac{a}{b} = \dfrac{1}{4}\]
So, \[a:b = 1:4\]
Since we got our desired answer we will don’t need to equate the second factor.

So, option A is correct.

Note: Students many times make mistake of writing the arithmetic mean of two numbers as \[\dfrac{{a + b}}{2}\] without looking at the numbers which are given to us as \[\dfrac{1}{a}\] and \[\dfrac{1}{b}\]. Also, many students try to find the factors of the equation using the method to find roots of a quadratic equation which is wrong.