Fun with Maths

Hint: Use the fact that polynomial equation formed by replacing x with 1/x i.e x-->1/x will have roots in AP.From there it easy considering roots as
a-d,a,a+d then applying vieta to get 'a' (which is a root of the polynomial obtained after transformation and must satistisy it)

Let the roots be $\dfrac{1}{a}$ , $\dfrac{1}{a+d}$ and $\dfrac{1}{a+2d}$ , then by Vieta's formulas:

$\begin{cases} \dfrac{1}{a} + \dfrac{1}{a+d} + \dfrac{1}{a+2d} = \dfrac{c}{10} & ...(1) \\ \dfrac{1}{a(a+d)} + \dfrac{1}{(a+d)(a+2d)} + \dfrac{1}{a(a+2d)} = - \dfrac{27}{5} & ...(2) \\ \color{#3D99F6}{\dfrac{1}{a(a+d)(a+2d)} = \dfrac{27}{10}} & \color{#3D99F6}{...(3)} \end{cases}$

$\begin{aligned} (2): \quad \dfrac{1}{a(a+d)} + \dfrac{1}{(a+d)(a+2d)} + \dfrac{1}{a(a+2d)} & = - \dfrac{27}{5} \\ \Rightarrow \frac{a+2d+a+a+d}{\color{#3D99F6}{a(a+d)(a+2d)}} & = - \frac{27}{5} \\ \dfrac{3(a+d)}{\color{#3D99F6}{a(a+d)(a+2d)}} & = - \dfrac{27}{5} \\ 3(a+d)\left(\color{#3D99F6}{\frac{27}{10}}\right) & = - \frac{27}{5} \\ \Rightarrow \color{#D61F06}{a+d} & \color{#D61F06}{= - \frac{2}{3}} \end{aligned}$

$\begin{aligned} (3): \quad \frac{1}{a(a+d)(a+2d)} & = \frac{27}{10} \\ a(\color{#D61F06}{a+d})(a+2d) & = \frac{10}{27} \\ a\left(\color{#D61F06}{- \frac{2}{3}}\right)(a+2d) & = \frac{10}{27} \\ \Rightarrow \color{#20A900}{a(a+2d)} & \color{#20A900}{= - \frac{5}{9}} \end{aligned}$

$\begin{aligned} (1): \quad \quad \quad \quad \quad \quad \quad \quad \frac{1}{a} + \frac{1}{a+d} + \frac{1}{a+2d} & = \frac{c}{10} \\ \frac{(a+d)(a+2d)+a(a+2d)+a(a+d)}{\color{#3D99F6}{a(a+d)(a+2d)}} & = \frac{c}{10} \\ \frac{(a+d)(a+2d+a)+a(a+2d)}{\color{#3D99F6}{a(a+d)(a+2d)}} & = \frac{c}{10} \\ \frac{2(\color{#D61F06}{a+d})^2+\color{#20A900}{a(a+2d)}}{\color{#3D99F6}{a(a+d)(a+2d)}} & = \frac{c}{10} \\ \left[ 2\left(\color{#D61F06}{ - \frac{2}{3}} \right)^2 \color{#20A900}{- \frac{5}{9}} \right]\left(\color{#3D99F6}{\frac{27}{10}}\right) & = \frac{c}{10} \\ \left[ \frac{8}{9} - \frac{5}{9} \right]\left(\color{#3D99F6}{\frac{27}{10}}\right) & = \frac{c}{10} \\ \frac{1}{3} \dot{} \left(\color{#3D99F6}{\frac{27}{10}}\right) & = \frac{c}{10} \\ \Rightarrow c & = \boxed{9} \end{aligned}$