Answer

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.

**Complete step by step 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.

Recently Updated Pages

What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main

If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main

The area of square inscribed in a circle of diameter class 10 maths JEE_Main

Common roots of the equations2sin 2x + sin 22x 2and class 10 maths JEE_Main

If alpha and beta are the roots of the equation x2 class 10 maths JEE_Main

The mean of discrete observations y1y2yn is given by class 10 maths JEE_Main

Other Pages

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Which of the following distance time graph is representing class 11 physics JEE_Main