Question

If A, G and 4 are AM, GM and HM of two numbers respectively and $2A+{{G}^{2}}=27$ then the numbers are
\[\begin{align}
  & \text{A}.\text{ 8},\text{ 2} \\
 & \text{B}.\text{ 8},\text{ 6} \\
 & \text{C}.\text{ 6},\text{ 3} \\
 & \text{D}.\text{ 6},\text{ 4} \\
\end{align}\]

Answer 1

Hint: We will use the formula of AM and GM and HM of two numbers which are given as \[AM=\dfrac{a+b}{2},GM=\sqrt{ab}\text{ and HM=}\dfrac{2ab}{a+b}\]. We will also use the fact that AM and GM are related as \[AM=\dfrac{a+b}{2}\text{ and }{{\left( GM \right)}^{2}}=ab\]. First, we will start by using HM = 4 and formula of HM. We will then try forming new equations using all the above formulas and relations and apply the relation given as $2A+{{G}^{2}}=27$. We will get a set of equations, which on solving will get us values of a and b.

Complete step-by-step answer:
Given that, A, G and 4 are AM, GM and HM of two numbers respectively.
Let a and b be two numbers then, we will use the formula of AM and GM and HM of two numbers which are given as
\[AM=\dfrac{a+b}{2},GM=\sqrt{ab}\text{ and HM=}\dfrac{2ab}{a+b}\]
Now, given that HM = 4 and
\[\begin{align}
  & \text{HM=}\dfrac{2ab}{a+b} \\
 & \Rightarrow 4=\dfrac{2ab}{a+b} \\
 & \text{Take }\dfrac{a+b}{2}=AM=A \\
 & \text{and ab=}{{\left( GM \right)}^{2}}={{G}^{2}} \\
\end{align}\]
Substituting the values in above equation we get:
\[\begin{align}
  & 4=\dfrac{{{G}^{2}}}{A} \\
 & \Rightarrow {{G}^{2}}=4A\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)} \\
\end{align}\]
Given that, \[2A+{{G}^{2}}=27\]
Substituting values of ${{G}^{2}}=4A$ from equation (i) we get:
\[\begin{align}
  & 2A+4A=27 \\
 & \Rightarrow 6A=27 \\
\end{align}\]
Divide by 6 we get:
\[\begin{align}
  & A=\dfrac{27}{6}=\dfrac{9}{2} \\
 & \Rightarrow A=\dfrac{9}{2} \\
\end{align}\]
From (i) we have ${{G}^{2}}=4A$
\[\begin{align}
  & {{G}^{2}}=4\left( \dfrac{9}{2} \right) \\
 & {{G}^{2}}=18 \\
\end{align}\]
So, we have obtained $AM;\text{ }A=\dfrac{9}{2}$
\[\begin{align}
  & \Rightarrow \dfrac{a+b}{2}=\dfrac{9}{2} \\
 & \Rightarrow a+b=9 \\
 & \Rightarrow b=9-a\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\
\end{align}\]
Also, we have obtained $GM;\text{ G=}\sqrt{18}$
\[\Rightarrow \sqrt{ab}=\sqrt{18}\]
Squaring both sides
\[\left( ab \right)=18\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}\]
Substituting value of b = 9-a from equation (ii) into equation (iii) we get
\[\begin{align}
  & \Rightarrow \left( a\left( 9-a \right) \right)=18 \\
 & \Rightarrow \left( 9a-{{a}^{2}} \right)=18 \\
 & \Rightarrow {{a}^{2}}-9a+18=0 \\
\end{align}\]
Splitting middle term, we get:
\[\begin{align}
  & {{a}^{2}}-6a-3a+18=0 \\
 & a\left( a-6 \right)-3\left( a-6 \right)=0 \\
 & a=3,6 \\
\end{align}\]
When a = 3 using equation (i),
\[\Rightarrow b=9-3=6\]
When a = 6 using equation (i),
\[\Rightarrow b=9-6=3\]
Therefore, the numbers are 3 and 6.

So, the correct answer is “Option C”.

Note: The key point to note in this question is that, when value of a is obtained to be 6 and 3 then, this is taken as answer. This step is wrong even if the value of both numbers are coming to be 3 and 6 then also we would consider b also separately to get the result.