Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Find two numbers whose arithmetic mean is 34 and the geometric mean is 16.

seo-qna
Last updated date: 20th Jun 2024
Total views: 404.1k
Views today: 13.04k
Answer
VerifiedVerified
404.1k+ views
Hint: Here, we need to find the two numbers. We will first assume the two numbers to be \[x\] and \[y\]. We will use the formula for arithmetic mean and geometric mean and simplify the equations. Then, we will use the formula for the square of the sum and difference of the square root of the two numbers to get two equations in terms of \[x\] and \[y\]. Finally, we will solve these two equations to get the values of \[x\] and \[y\], and hence, find the two numbers.

Formula Used: We will use the following formulas to solve the question:
The arithmetic mean of two numbers \[a\] and \[b\] is given by the formula \[A.M. = \dfrac{{a + b}}{2}\].
The geometric mean of two numbers \[a\] and \[b\] is given by the formula \[G.M. = \sqrt {ab} \].
The square of the sum of two numbers \[a\] and \[b\] is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
The square of the difference of two numbers \[a\] and \[b\] is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].

Complete step-by-step answer:
Let the two numbers be \[x\] and \[y\].
We will now find the arithmetic mean using the formula.
Substituting \[a = x\] and \[b = y\] in the formula \[A.M. = \dfrac{{a + b}}{2}\], we get
\[A.M. = \dfrac{{x + y}}{2}\]
It is given that the arithmetic mean is 34.
Therefore, we get
\[ \Rightarrow 34 = \dfrac{{x + y}}{2}\]
Multiplying both sides by 2, we get
\[\begin{array}{l} \Rightarrow 34 \times 2 = \left( {\dfrac{{x + y}}{2}} \right) \times 2\\ \Rightarrow x + y = 68\end{array}\]
We will now find the geometric mean using the formula.
Substituting \[a = x\] and \[b = y\] in the formula \[G.M. = \sqrt {ab} \], we get
\[G.M. = \sqrt {xy} \]
It is given that the geometric mean is 16.
Therefore, we get
\[16 = \sqrt {xy} \]
Now, we will use the algebraic identities for the square of sum and difference of two numbers to find two equations in terms of \[x\] and \[y\].
Substituting \[a = \sqrt x \] and \[b = \sqrt y \] in the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[ \Rightarrow {\left( {\sqrt x + \sqrt y } \right)^2} = {\left( {\sqrt x } \right)^2} + {\left( {\sqrt y } \right)^2} + 2\left( {\sqrt x } \right)\left( {\sqrt y } \right)\]
Simplifying the expression, we get
\[ \Rightarrow {\left( {\sqrt x + \sqrt y } \right)^2} = x + y + 2\sqrt {xy} \]
Substituting \[x + y = 68\] and \[\sqrt {xy} = 16\] in the equation, we get
\[ \Rightarrow {\left( {\sqrt x + \sqrt y } \right)^2} = 68 + 2 \times 16\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {\left( {\sqrt x + \sqrt y } \right)^2} = 68 + 32\\ \Rightarrow {\left( {\sqrt x + \sqrt y } \right)^2} = 100\end{array}\]
Taking square root on both sides, we get
\[ \Rightarrow \sqrt x + \sqrt y = 10 \ldots \ldots \ldots \left( 1 \right)\]
Substituting \[a = \sqrt x \] and \[b = \sqrt y \] in the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], we get
\[ \Rightarrow {\left( {\sqrt x - \sqrt y } \right)^2} = {\left( {\sqrt x } \right)^2} + {\left( {\sqrt y } \right)^2} - 2\left( {\sqrt x } \right)\left( {\sqrt y } \right)\]
Simplifying the expression, we get
\[ \Rightarrow {\left( {\sqrt x - \sqrt y } \right)^2} = x + y - 2\sqrt {xy} \]
Substituting \[x + y = 68\] and \[\sqrt {xy} = 16\] in the equation, we get
\[ \Rightarrow {\left( {\sqrt x - \sqrt y } \right)^2} = 68 - 2 \times 16\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow {\left( {\sqrt x - \sqrt y } \right)^2} = 68 - 32\\ \Rightarrow {\left( {\sqrt x - \sqrt y } \right)^2} = 36\end{array}\]
Taking square root on both sides, we get
\[ \Rightarrow \sqrt x - \sqrt y = 6 \ldots \ldots \ldots \left( 2 \right)\]
We will solve the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] to find the value of \[x\] and \[y\].
Adding equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\], we get
\[\begin{array}{l}{\rm{ }}\sqrt x + \sqrt y = 10\\\underline {{\rm{ }}\sqrt x - \sqrt y = {\rm{ }}6} \\2\sqrt x {\rm{ }} = 16\end{array}\]
Dividing both sides of the resulting equation by 2, we get
\[\begin{array}{l} \Rightarrow \dfrac{{2\sqrt x }}{2} = \dfrac{{16}}{2}\\ \Rightarrow \sqrt x = 8\end{array}\]
Squaring both sides, we get
\[ \Rightarrow x = 64\]
Substituting \[\sqrt x = 8\] in \[\sqrt x + \sqrt y = 10\], we get
\[ \Rightarrow 8 + \sqrt y = 10\]
Subtracting 8 from both sides, we get
\[ \Rightarrow \sqrt y = 2\]
Squaring both sides, we get
\[ \Rightarrow y = 4\]
\[\therefore\] The two numbers are 64 and 4.

Note: We can verify your answer by finding the arithmetic mean and geometric mean of 64 and 4.
Substituting \[a = 64\] and \[b = 4\] in the formula for arithmetic mean, we get
\[\begin{array}{l} \Rightarrow A.M. = \dfrac{{64 + 4}}{2}\\ \Rightarrow A.M. = \dfrac{{68}}{2}\\ \Rightarrow A.M. = 34\end{array}\]
Substituting \[a = 64\] and \[b = 4\] in the formula for geometric mean, we get
\[\begin{array}{l} \Rightarrow G.M. = \sqrt {64 \times 4} \\ \Rightarrow G.M. = \sqrt {256} \\ \Rightarrow G.M. = 16\end{array}\]
Hence, we have verified the answer.