For the two positive numbers a , b, if a , b | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$a , b , \frac{1}{18} \rightarrow GP$

$\frac{ a }{18}= b ^2$

$\frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP$

$\frac{1}{ a }+\frac{1}{ b }=2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$a , b , \frac{1}{18} ightarrow GP$

$\frac{ a }{18}= b ^2$

$\frac{1}{ a }, 10, \frac{1}{ b } ightarrow AP$

$\frac{1}{ a }+\frac{1}{ b }=20$

$\Rightarrow a + b =20 ab , 	ext { from eq. (i) } ; 	ext { we get }$

$\Rightarrow 18 b ^2+ b =360 b ^3$

$\Rightarrow 360 b ^2-18 b -1=0 \quad\{\because b 
eq 0\}$

$\Rightarrow b =\frac{18 \pm \sqrt{324+1440}}{720}$

$\Rightarrow b=\frac{18+\sqrt{1764}}{720} \quad\{\because b > 0\}$

$\Rightarrow b=\frac{1}{12}$

$\Rightarrow a=18 	imes \frac{1}{144}=\frac{1}{8}$

Now, $16 a+12 b=16 	imes \frac{1}{8}+12 	imes \frac{1}{12}=3$

For the two positive numbers $a , b$, if $a , b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{ a }, 10$ and $\frac{1}{ b }$ are in an arithmetic progression, then, $16 a+12 b$ is equal to $.........$.

Similar Questions

The arithmetic mean and the geometric mean of two distinct 2-digit numbers $x$ and $y$ are two integers one of which can be obtained by reversing the digits of the other (in base 10 representation). Then, $x+y$ equals

If $a_1, a_2...,a_n$ an are positive real numbers whose product is a fixed number $c$ , then the minimum value of $a_1 + a_2 +.... + a_{n - 1} + 2a_n$ is

${2^{\sin \theta }} + {2^{\cos \theta }}$ is greater than

If the arithmetic, geometric and harmonic means between two positive real numbers be $A,\;G$ and $H$, then

Let $x, y, z$ be three non-negative integers such that $x+y+z=10$. The maximum possible value of $x y z+x y+y z+z x$ is