Question

If A, G and 4 are AM, GM and HM of two numbers respectively and $2A+{{G}^{2}}=27$ then the numbers are

& \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{aligned}$$

Answer 1

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

& \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{aligned}$$ Substituting the values in above equation we get: $$\begin{aligned} & 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{aligned}$$ Given that, $$2A+{{G}^{2}}=27$$ Substituting values of ${{G}^{2}}=4A$ from equation (i) we get: $$\begin{aligned} & 2A+4A=27 \\\ & \Rightarrow 6A=27 \\\ \end{aligned}$$ Divide by 6 we get: $$\begin{aligned} & A=\dfrac{27}{6}=\dfrac{9}{2} \\\ & \Rightarrow A=\dfrac{9}{2} \\\ \end{aligned}$$ From (i) we have ${{G}^{2}}=4A$ $$\begin{aligned} & {{G}^{2}}=4\left( \dfrac{9}{2} \right) \\\ & {{G}^{2}}=18 \\\ \end{aligned}$$ So, we have obtained $AM;\text{ }A=\dfrac{9}{2}$ $$\begin{aligned} & \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{aligned}$$ Also, we have obtained $GM;\text{ G=}\sqrt{18}$ $$\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{aligned} & \Rightarrow \left( a\left( 9-a \right) \right)=18 \\\ & \Rightarrow \left( 9a-{{a}^{2}} \right)=18 \\\ & \Rightarrow {{a}^{2}}-9a+18=0 \\\ \end{aligned}$$ Splitting middle term, we get: $$\begin{aligned} & {{a}^{2}}-6a-3a+18=0 \\\ & a\left( a-6 \right)-3\left( a-6 \right)=0 \\\ & a=3,6 \\\ \end{aligned}$$ When a = 3 using equation (i), $$\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.