For the two positive numbers a , b, if a , b and 1 18 are in a ge… - Vidyadip

MCQ

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 $.........$.

✓
$3$
B
$2$
C
$1$
D
$0$

✓

Answer

Correct option: A.

$3$

a
$a , b , \frac{1}{18} \rightarrow GP$

$\frac{ a }{18}= b ^2$

$\frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP$

$\frac{1}{ a }+\frac{1}{ b }=20$

$\Rightarrow a + b =20 ab , \text { from eq. (i) } ; \text { we get }$

$\Rightarrow 18 b ^2+ b =360 b ^3$

$\Rightarrow 360 b ^2-18 b -1=0 \quad\{\because b \neq 0\}$

$\Rightarrow b =\frac{18 \pm \sqrt{324+1440}}{720}$

$\Rightarrow b=\frac{18+\sqrt{1764}}{720} \quad\{\because b > 0\}$

$\Rightarrow b=\frac{1}{12}$

$\Rightarrow a=18 \times \frac{1}{144}=\frac{1}{8}$

Now, $16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3$

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