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 $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ 5& 4& a_6\\ a_7& a_8& a_9\end{array}\right|$ can be expressed in the lowest form as $\frac{p}{q}$, find $(p+q)$

Answer 1

Answer: 71

$a_3=\frac{1}{\frac{1}{5}-\frac{1}{20}}=\frac{20}{3}$
$a_2=\frac{1}{\frac{3}{20}-\frac{1}{20}}=10$
$a_1=\frac{1}{\frac{1}{10}-\frac{1}{20}}=20$
$a_6=\frac{1}{\frac{1}{4}+\frac{1}{20}}=\frac{10}{3}$
$a_7=\frac{1}{\frac{3}{10}+\frac{1}{20}}=\frac{20}{7}$
$a_8=\frac{1}{\frac{7}{20}+\frac{1}{20}}=\frac{5}{2}$
$a_9=\frac{1}{\frac{2}{5}+\frac{1}{20}}=\frac{20}{9}$
$20^3\left|\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{3}\\ \frac{1}{4}& \frac{1}{5}& \frac{1}{6}\\ \frac{1}{7}& \frac{1}{8}& \frac{1}{9}\end{array}\right|=\frac{50}{21}$
$\Rightarrow p+q=71$