Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If H.M. between two numbers is $\dfrac{{16}}{5}$. Their A.M. is A, G.M. is G and $2A + {\text{ }}{G^2} = 26$. Then find the two numbers.

Last updated date: 20th Jun 2024
Total views: 54.3k
Views today: 1.54k
Verified
54.3k+ views
Hint: In this question, use the formula of relation between the geometric mean, harmonic mean and arithmetic mean. Then, you will get the two equations, solve it and you will get the answer.

In the question, it is given that,
Harmonic mean of two numbers is $\dfrac{{16}}{5}$.
The relation between arithmetic mean and geometric mean is $2A + {\text{ }}{G^2} = 26$ .
We have to find the two numbers using the given conditions.
We know that, $\dfrac{{G{M^2}}}{{AM}} = HM$
Now it is given that, $2A + {\text{ }}{G^2} = 26$
${G^2} = {\text{ }}26 - {\text{ }}2A$
By putting in the formula,
$\dfrac{{{{(26 - 2A)}^{}}}}{A} = \dfrac{{16}}{5}$
By cross multiplication,
$5{\left( {26 - 2A} \right)^{}} = {\text{ }}16A$
$130{\text{ }}-{\text{ }}10A{\text{ }} = {\text{ }}16A$
We are solving this equation
$130{\text{ }} = {\text{ }}16A{\text{ }} + 10A$
$130{\text{ }} = {\text{ }}26A$
$A = \dfrac{{130}}{{26}}$
$A = 5$
Therefore, arithmetic mean $= 5$
Geometric mean = ${G^2} = {\text{ }}26{\text{ }}-{\text{ }}2A$
$G = \sqrt {26 - 2A}$
Substituting the value of A in the equation, we will get
$= \sqrt {26 - 2 \times 5}$
$= \sqrt {26 - 10}$
$= \sqrt {16}$
By taking the square root we will get the value of G
$G{\text{ }} = 4$
Geometric mean = $4$
Now, let say the two required numbers be a and b
If arithmetic mean = 5
This implies that, $\dfrac{{a + b}}{2} = 5$
$a{\text{ }} + {\text{ }}b = 10\;$
If Geometric mean = 4
This implies that $\sqrt {ab} = 4$
Therefore, $ab{\text{ }} = {\text{ }}{4^2}$
$ab{\text{ }} = {\text{ }}16$
Now, we have two equations,
$a{\text{ }} + {\text{ }}b{\text{ }} = {\text{ }}10$
$ab{\text{ }} = {\text{ }}16$
solving the two equations,
$a = 10-b$
substituting this value in equation 2
$\left( {10{\text{ }}-{\text{ }}b} \right){\text{ }}b = 16$
$10b-{b^2} = 16$
${b^2}-10b + 16 = 0$
Solving the question by middle term splitting
${b^2}-{\text{ }}8b{\text{ }}-{\text{ }}2b{\text{ }} + {\text{ }}16{\text{ }} = {\text{ }}0$
$b{\text{ }}\left( {b{\text{ }}-{\text{ }}8} \right){\text{ }}-{\text{ }}2{\text{ }}\left( {b{\text{ }} - {\text{ }}8} \right){\text{ }} = {\text{ }}0$
$\left( {b{\text{ }}-{\text{ }}8} \right){\text{ }}\left( {b{\text{ }}-{\text{ }}2} \right){\text{ }} = {\text{ }}0$
Therefore, $b{\text{ }} = {\text{ }}8{\text{ }}and{\text{ }}b = {\text{ }}2$
Therefore $a{\text{ }} = {\text{ }}2{\text{ }}or{\text{ }}a{\text{ }} = {\text{ }}8$

Therefore, the required two numbers are $2{\text{ }}and{\text{ }}8$.

Note: The two numbers are found using the middle term splitting. Both a and b got the same values, So, if we take a = 8, then b = 2 and if we take a = 2, then b = 8.

‌
‌