Question

The AM of two given positive numbers is 2. If the larger number is increased by 1, the GM of the number becomes equal to the AM of the given numbers. Then the HM of the given numbers is
A.$\dfrac{3}{2}$
B.$\dfrac{2}{3}$
C.$\dfrac{1}{2}$
D.None of these

Answer 1

Hint: Let two numbers be $a$ and $b$. Write the AM of the two numbers and equate it to 2 to form an equation. Similarly, write the GM of the two numbers such that the larger number is increased by 1 and equate it to 2. Solve the equations to find the numbers and then find the HM of the numbers.

Complete step-by-step answer:
First of all, we will let the two numbers as $a$ and $b$, such that $b$ is larger than $a$
We know that the arithmetic mean of two numbers is given by $\dfrac{{a + b}}{2}$.
We are given that the AM of two numbers is 2. Hence, we can write it as,
$\dfrac{{a + b}}{2} = 2 \\\ a + b = 4{\text{ }}\left( 1 \right) \\\$
According to the given number, if the larger number is increased by 1, the GM of the numbers become equal to the AM of the given numbers.
If $m$ and $n$ are any two numbers, then the GM is given by $\sqrt {mn}$.
Hence, GM of $a$ and $b + 1$ is given by $\sqrt {a\left( {b + 1} \right)}$, equating it to the AM of the given numbers.
$\sqrt {a\left( {b + 1} \right)} = 2 \\\ a\left( {b + 1} \right) = 4{\text{ }}\left( 2 \right) \\\$
Solve for the values of $a$ and $b$ by equating (1) and (2).
$a + b = a\left( {b + 1} \right) \\\ a + b = ab + a \\\ a = 1 \\\$
Hence,
$1 + b = 4 \\\ b = 3 \\\$
Then numbers are 1 and 3.
H.M of two numbers $m$ and $n$ is given by $\dfrac{{2mn}}{{m + n}}$
Then, the HM of 1 and 3 is, $\dfrac{{2\left( 1 \right)\left( 3 \right)}}{{1 + 3}} = \dfrac{6}{4} = \dfrac{3}{2}$
Hence, option A is correct.

Note: AM stands for arithmetic mean and is obtained by adding all the given values and then dividing it by the total number of values. Also, GM stands for geometric mean and is obtained by taking the ${n^{th}}$ root of the product of $n$ terms. HM stands for harmonic mean and is calculated as the reciprocal of arithmetic mean.