Let frac{1}{16}, a and b be in G.P. and frac{ | vedclass

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(C) Since $\frac{1}{16}, a, b$ are in $G.P.$,we have $a^2 = \frac{b}{16}$,which implies $b = 16a^2$.Since $\frac{1}{a}, \frac{1}{b}, 6$ are in $A.P.$,

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$14$

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(C) Since $\frac{1}{16}, a, b$ are in $G.P.$,we have $a^2 = \frac{b}{16}$,which implies $b = 16a^2$.Since $\frac{1}{a}, \frac{1}{b}, 6$ are in $A.P.$,we have $\frac{2}{b} = \frac{1}{a} + 6$.Substituting $b = 16a^2$ into the $A.P.$ equation:$\frac{2}{16a^2} = \frac{1}{a} + 6$$\frac{1}{8a^2} = \frac{1}{a} + 6$Multiply by $8a^2$:$1 = 8a + 48a^2$$48a^2 + 8a - 1 = 0$$(12a - 1)(4a + 1) = 0$Since $a > 0$,we have $a = \frac{1}{12}$.Then $b = 16 	imes (\frac{1}{12})^2 = 16 	imes \frac{1}{144} = \frac{1}{9}$.Finally,$72(a + b) = 72(\frac{1}{12} + \frac{1}{9}) = 72(\frac{3 + 4}{36}) = 72(\frac{7}{36}) = 2 	imes 7 = 14$.

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

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