Question

What is the geometric mean of 275 and 11?

Answer 1

Hint: We first form the condition for $a$ to be the geometric mean of 275 and 11. We have to find the square root of the multiplication of the given numbers. We use prime factorisation of the multiplication to find the solution.

Complete step by step solution:
We have to find the geometric mean of 275 and 11.
This means if $a$ is the geometric mean of 275 and 11 then we have $\dfrac{275}{a}=\dfrac{a}{11}$ which gives ${{a}^{2}}=11\times 275$.
To find the geometric mean we need to find the square root of $11\times 275$.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 2 we get \[{{a}^{\dfrac{1}{2}}}=\sqrt[2]{11\times 275}\].
We only need to find the prime factorisation of the given number 275 as 11 is a prime number.
$\begin{align}
  & 5\left| \!{\underline {\,
  27 \,}} \right. 5 \\
 & 5\left| \!{\underline {\,
  55 \,}} \right. \\
 & 11\left| \!{\underline {\,
  11 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[275=5\times 5\times 11\].
For finding the square root, we need to take one digit out of the two same number of primes.
This means in the square root value of \[275\times 11=5\times 5\times 11\times 11\], we will take out one 5 and one 11 from the multiplication.
So, $\sqrt[2]{11\times 275}=\sqrt[2]{5\times 5\times 11\times 11}=5\times 11=55$.
Therefore, the geometric mean of 275 and 11 is 55.

Note: We can also use the variable from where we can take \[x=\sqrt[2]{11\times 275}\]. But we need to remember that we can’t use the square on both sides of the equation \[x=\sqrt[2]{11\times 275}\] as in that case we are taking one extra value as a root value. We can also verify that $\dfrac{275}{55}=\dfrac{55}{11}=5$.