For two positive numbers a, b, if a, b, 118 are in a geometric progression, while 1a, 10, …

Question

For two positive numbers $a, b$, if $a, b, \dfrac{1}{18}$ are in a geometric progression, while $\dfrac1a, 10, \dfrac1b$ are in an arithmetic progression, then $16a+12b$ is equal to \_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

G.P.: $b^2 = \dfrac{a}{18} \Rightarrow a = 18b^2$. A.P.: $\dfrac1a+\dfrac1b = 20$.

Substituting $a=18b^2$: $\dfrac{1}{18b^2}+\dfrac1b=20 \implies 1+18b = 360b^3 \implies 360b^3-18b-1=0.$

Solving (with $b>0$) gives $b=\dfrac{1}{12}$, then $a=18\left(\dfrac{1}{144}\right)=\dfrac18$.

$$16a+12b = 16\left(\frac18\right)+12\left(\frac{1}{12}\right) = 2+1 = 3.$$

Solutions · 1