Question

If $a_{1}, a_{2}, a_{3}, 5,4, a_{6}, a_{7}, a_{8}, a_{9}$ are in $H.P$., then the value of the determinant $\begin{vmatrix}a_{1} & a_{2} & a_{3} \\ 5 & 4 & a_{6} \\ a_{7} & a_{8} & a_{9}\end{vmatrix}$ can be expressed in the lowest form as $\frac{p}{q}$, find $(p +q)$

Answer 1

$a_{3}=\frac{1}{\frac{1}{5}-\frac{1}{20}}=\frac{20}{3}$
$a_{2}=\frac{1}{\frac{3}{20}-\frac{1}{20}}=10$
$a_{1}=\frac{1}{\frac{1}{10}-\frac{1}{20}}=20$
$a_{6}=\frac{1}{\frac{1}{4}+\frac{1}{20}}=\frac{10}{3}$
$a_{7}=\frac{1}{\frac{3}{10}+\frac{1}{20}}=\frac{20}{7}$
$a_{8}=\frac{1}{\frac{7}{20}+\frac{1}{20}}=\frac{5}{2}$
$a_{9}=\frac{1}{\frac{2}{5}+\frac{1}{20}}=\frac{20}{9}$
$20^{3}\begin{vmatrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} \\ \frac{1}{7} & \frac{1}{8} & \frac{1}{9}\end{vmatrix}=\frac{50}{21}$
$\Rightarrow p+ q=71$