JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 23)
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 $$16a+12b$$ is equal to _________.
Answer
3
Explanation
$$
\begin{aligned}
& \mathrm{a}, \mathrm{b}, \frac{1}{18} \rightarrow \mathrm{GP} \\\\
& \frac{\mathrm{a}}{18}=\mathrm{b}^2\quad...(i) \\\\
& \frac{1}{\mathrm{a}}, 10, \frac{1}{\mathrm{~b}} \rightarrow \mathrm{AP} \\\\
& \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=20 \\\\
& \Rightarrow \mathrm{a}+\mathrm{b}=20 \mathrm{ab}, \text { from eq. (i) } ; \text { we get } \\\\
& \Rightarrow 18 \mathrm{~b}^2+\mathrm{b}=360 \mathrm{~b}^3 \\\\
& \Rightarrow 360 \mathrm{~b}^2-18 \mathrm{~b}-1=0 \quad\{\because \mathrm{b} \neq 0\} \\\\
& \Rightarrow \mathrm{b}=\frac{18 \pm \sqrt{324+1440}}{720} \\\\
& \Rightarrow b =\frac{18+\sqrt{1764}}{720} \quad\{\because \mathrm{b}>0\} \\\\
& \Rightarrow b = \frac{1}{12} \\\\
& \Rightarrow \mathrm{a}=18 \times \frac{1}{144}=\frac{1}{8} \\\\
& \text{Now},16\mathrm{a}+12 \mathrm{~b}=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3
\end{aligned}
$$
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