Let 116, a, b be in G.P.

Question

Let $\dfrac{1}{16}, a, b$ be in G.P. and $\dfrac1a, \dfrac1b, 6$ be in A.P., where $a,b>0$. Then $72(a+b)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 14

Accepted Answer

G.P.: $a^2 = \dfrac{b}{16} \Rightarrow b=16a^2$. A.P.: $\dfrac2b = \dfrac1a+6$.

Substituting: $\dfrac{2}{16a^2} = \dfrac{1+6a}{a} \Rightarrow 1 = 8a(1+6a) \Rightarrow 48a^2+8a-1=0 \Rightarrow (4a+1)(12a-1)=0.$

Since $a>0$, $a=\dfrac{1}{12}$, and $b=16\left(\dfrac{1}{144}\right)=\dfrac19$.

$$72(a+b) = 72\left(\frac{1}{12}+\frac19\right) = 6+8 = 14.$$

Solutions · 1