JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Sequences and Series

From the geometric progression $a,b,\frac{1}{18}$ the middle term gives $b^2=\frac{a}{18}$, so $a=18b^2$. From the arithmetic progression $\frac{1}{a},10,\frac{1}{b}$ the middle term gives $\frac{1}{a}+\frac{1}{b}=20$. Substituting $a=18b^2$ leads to $360b^2-18b-1=0$, whose positive root is $b=\frac{1}{12}$, and then $a=\frac{1}{8}$. Therefore $16a+12b=16\cdot \frac{1}{8}+12\cdot \frac{1}{12}=2+1=3$.