Question

Let $\frac{1}{16},a$ and $b$ be in G.P. and $\frac{1}{a},\frac{1}{b},6$ be in A.P., where $a,b,>0$. Then $72(a+b)$ is equal to

Answer 1

Step 1: Finding the value of $a$ and $b$

Given, $\frac{1}{16},a$ and $b$are in G.P, hence their geometric mean will be equal to,

$$\begin{gathered} a^2=\frac{b}{16} \\ \Rightarrow b=16a^2\dots \dots \left(1\right) \end{gathered}$$

It is also given that, $\frac{1}{a},\frac{1}{b}$ and $6$are in AP, hence their arithmetic mean will be equal to,

$\frac{2}{b}=\frac{1}{a}+6\dots \dots \left(2\right)$

Substituting the value of $b$from equation $\left(1\right)$ in equation $\left(2\right)$ we get,

$$\begin{gathered} \frac{2}{16a^2}=\frac{1}{a}+6 \\ \Rightarrow \frac{1}{8a^2}=\frac{1+6a}{a} \\ \Rightarrow 8a+48a^2=1 \\ \Rightarrow 48a^2+8a^2-1=0 \\ \Rightarrow 48a^2+12a-4a-1=0 \\ \Rightarrow 12a\left(4a+1\right)-\left(4a+1\right)=0 \\ \Rightarrow \left(12a-1\right)\left(4a+1\right)=0 \\ \Rightarrow a=\frac{1}{12},-\frac{1}{4} \end{gathered}$$

Here the value of $a$ will be equal to $\frac{1}{12}$ because $a>0$. Therefore the value of $b$ will be equal to,

$$\begin{gathered} b=16a^2 \\ \Rightarrow b=16\times {\left(\frac{1}{12}\right)}^2 \\ \Rightarrow b=\frac{16}{144} \end{gathered}$$

Step 2: Calculating the value of $72\left(a+b\right)$

Therefore the value of the given expression is equal to,

$$\begin{gathered} 72\left(a+b\right)=72\left(\frac{1}{12}+\frac{16}{144}\right) \\ =72\left(\frac{12+16}{144}\right) \\ =72\left(\frac{28}{144}\right) \\ =14 \end{gathered}$$

Hence, is the value of the given expression $72\left(a+b\right)=$$14$.