Question

If the A.M. of two numbers exceeds their G.M. by 10 and their H.M. by 16, find the numbers.

Answer 1

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.