If the first three terms of the sequence frac | vedclass

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(C) If $\frac{1}{16}, a, b$ are in $G.P.$,then $a^2 = \frac{b}{16}$,which implies $b = 16a^2$ ..... $(i)$If $a, b, \frac{1}{6}$ are in $H.P.$,then $b

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Both $(a)$ and $(b)$ are true

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(C) If $\frac{1}{16}, a, b$ are in $G.P.$,then $a^2 = \frac{b}{16}$,which implies $b = 16a^2$ ..... $(i)$If $a, b, \frac{1}{6}$ are in $H.P.$,then $b = \frac{2 \cdot a \cdot \frac{1}{6}}{a + \frac{1}{6}} = \frac{2a}{6a + 1}$ ..... $(ii)$Equating $(i)$ and $(ii)$,we get $16a^2 = \frac{2a}{6a + 1}$.For $a 
eq 0$,$8a = \frac{1}{6a + 1}$,which gives $48a^2 + 8a - 1 = 0$.Factoring the quadratic equation: $(4a + 1)(12a - 1) = 0$.Thus,$a = -\frac{1}{4}$ or $a = \frac{1}{12}$.If $a = -\frac{1}{4}$,then $b = 16(-\frac{1}{4})^2 = 1$.If $a = \frac{1}{12}$,then $b = 16(\frac{1}{12})^2 = 16 \cdot \frac{1}{144} = \frac{1}{9}$.Therefore,both pairs are valid solutions.

If the first three terms of the sequence $\frac{1}{16}, a, b, \frac{1}{6}$ are in a geometric progression and the last three terms are in a harmonic progression,then the values of $a$ and $b$ are:

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