Question

The two numbers between $\frac{1}{16},\frac{1}{6}$ such that first three may be in G.P. and the last three in H.P. are respectively.

• A. $\frac{1}{12},\frac{1}{9}$
  
• B. $-\frac{1}{4},-1$
  
• C. $\frac{-1}{12},\frac{-1}{9}$
  
• D. $\frac{1}{4},1$

Answer 1

The correct option is A

$\frac{1}{12},\frac{1}{9}$

Let the two numbers be a, b. Then the set of numbers is $\frac{1}{16},a,b,\frac{1}{6}$.
Given: $\frac{1}{16},a,b$ are in G.P.
$a^2=\frac{1}{16}b$
$\Rightarrow b=16a^2$
Given: $a,b,\frac{1}{6}$ are in H.P. $b=\frac{2a\frac{1}{6}}{a+\frac{1}{6}}=\frac{2a}{6a+1}\Rightarrow 16a^2=\frac{2a}{6a+1}\Rightarrow 8a(6a+1)=148a^2+8a=1\Rightarrow 48a^2+8a-1=0\Rightarrow 48a^2+12a-4a-1=012a(4a+1)-1(4a+1)=0\Rightarrow (4a+1)(12a-1)=012a(4a+1)-1(4a+1)=0\Rightarrow (4a+1)(12a-1)=04a=-1\Rightarrow a=-\frac{1}{4}$
But a is a positive number. Hence $a\ne -\frac{1}{4}$
$12a-1\Rightarrow a=\frac{1}{12}$
$b=16{\left(\frac{1}{12}\right)}^2=16-\frac{1}{144}=\frac{1}{9}$