Answer

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**Hint:**First, find the relation between G.M. and H.M. Then, substitute the value of A.M. and H.M. in the formula ${G^2} = AH$ and simplify to get the value of $G$. After that use the formula $G = \sqrt {ab} $ and write $b$ in terms of $a$. Also, find the value of $A$ from the value of $G$. Then, use the formula $A = \dfrac{{a + b}}{2}$ and substitute the value of $b$ which is in terms of $a$. Now solve the quadratic equation formed. Substitute the value of $a$ in $b$ to get the value of $b$. The value derived from $a$ and $b$ is the desired results.

**Complete step by step answer:**

Given: A.M. of two numbers exceeds their G.M. by 10 and their H.M. by 16

Let the numbers be a and b, the A.M. of two numbers be A, the G.M. be G and the H.M. be H.

Now, the A.M. of two numbers exceeds their G.M. by 10,

$A = G + 10$ ……………..….. (1)

Also, the A.M. of two numbers exceeds their H.M. by 16,

$A = H + 16$ ……………….….. (2)

Equate both equations,

$G + 10 = H + 16$

Move 16 to the left side of the equation and subtract from 10,

$G - 6 = H$ ………………..….. (3)

As we know that the square of the geometric mean is equal to the product of arithmetic mean and harmonic mean.

${G^2} = AH$

Substitute the values of A and H from the equations (1) and (3),

${G^2} = \left( {G + 10} \right)\left( {G - 6} \right)$

Multiply the terms on the right side,

${G^2} = {G^2} + 10G - 6G - 60$

Cancel out ${G^2}$ from both sides and move the constant term to the other side,

$4G = 60$

Divide both sides by 4,

$G = 15$

As $G = \sqrt {ab} $substitute it in the above equation,

$\sqrt {ab} = 15$

Square both sides of the equation,

$ab = 225$

Find the value of $b$in terms of $a$,

$b = \dfrac{{225}}{a}$ ………………...….. (4)

Substitute the value of $G$ in equation (1),

$A = 15 + 10$,

As $A = \dfrac{{a + b}}{2}$. Then,

$\dfrac{{a + b}}{2} = 25$

Multiply both sides by 2,

$a + b = 50$

Substitute the value of $b$ from equation (4),

$a + \dfrac{{225}}{a} = 50$

Take LCM on the left side,

$\dfrac{{{a^2} + 225}}{a} = 50$

Multiply both sides by $a$,

${a^2} + 225 = 50a$

Move \[50a\] on the left side and factor it,

$ {a^2} - 50a + 225 = 0 \\

{a^2} - 45a - 5a + 225 = 0 \\

$

Take out common factors,

$

a\left( {a - 45} \right) - 5\left( {a - 45} \right) = 0 \\

\left( {a - 45} \right)\left( {a - 5} \right) = 0 \\

$

Set $\left( {a - 45} \right)$ to zero,

\[

a - 45 = 0 \\

a = 45 \\

\]

Substitute the value of $a$ in equation (4),

$

b = \dfrac{{225}}{{45}} \\

= 5 \\

$

Set $\left( {a - 5} \right)$ to zero,

\[

a - 5 = 0 \\

a = 5 \\

\]

Substitute the value of $a$ in equation (4),

$ b = \dfrac{{225}}{5} \\

= 45 \\

$

**Hence, the numbers are 5 and 45.**

**Note:**

An arithmetic sequence is a pattern of numbers in which the difference between consecutive terms of the sequence remains constant throughout the sequence.

A geometric progression is a sequence of numbers in which any two consecutive terms of the sequence have a common ratio.

Harmonic progression is the sequence that forms an arithmetic sequence when the reciprocal of terms is taken in order.

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