Question

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

Answer 1

Correct Answer: 3

From the given conditions:

\( \frac{a}{18} = b^2 \quad (1). \)

Also, since \(\frac{1}{a}, 10, \frac{1}{b}\) are in arithmetic progression:

\( \frac{1}{a} + \frac{1}{b} = 20. \)

Using equation (1) and simplifying:

\( a + b = 20ab. \)

Substituting \(b = \sqrt{\frac{a}{18}}\), we solve for \(a\) and \(b\), eventually leading to:

\( a = \frac{1}{8}, \quad b = \frac{1}{12}. \)

Now compute \(16a + 12b\):

\( 16a + 12b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 2 + 1 = 3. \)

Answer 2

The correct answer is 3.
$a,b,\frac{1}{18}\to GP$
$\frac{a}{18}=b^2....$(i)
$\frac{1}{a},10,\frac{1}{b}\to AP$
$\frac{1}{a}+\frac{1}{b}=20$
$\Rightarrow a+b=20ab\text{, from eq. (i); we get}$
$\Rightarrow 18b^2+b=360b^3$
$\Rightarrow 360b^2-18b-1=0\{\because b\ne 0\}$
$\Rightarrow b=\frac{18\pm \sqrt{324+1440}}{720}$
$\Rightarrow b=\frac{18+\sqrt{1764}}{720}\{\because b>0\}$
$\Rightarrow b=\frac{1}{12}$
$\Rightarrow a=18\times \frac{1}{144}=\frac{1}{8}$
Now, $16a+12b=16\times \frac{1}{8}+12\times \frac{1}{12}=3$