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

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.