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If $a_1,a_2,a_3,....,a_r$ are in G.P., then prove that the determinant $\begin{vmatrix} a_{r-1} & a_{r-5} & a_{r-9} \\ a_{r-7} & a_{r-11} & a_{r-15} \\ a_{r-13} & a_{r-17} & a_{r-21} \end{vmatrix} is\; independent\; of\; r$

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• If each element of a row (or a column) of a determinant is multiplied by constant k,and the added to the corresponding elements of some other row or column,then the value of the determinant remaining the same.
Let $\Delta=\begin{vmatrix} a_{r-1} & a_{r-5} & a_{r-9} \\ a_{r-7} & a_{r-11} & a_{r-15} \\ a_{r-13} & a_{r-17} & a_{r-21} \end{vmatrix}$
Let the ratio of the G.P be R
Therefore $\frac{a_r}{a_{r-1}}=R.$
$\Rightarrow \frac{a_{r-1}}{a_{r-5}}=R^4,\frac{a_{r-9}}{a_{r-1}}=R^8.$
Multiply $C_2$ by $R^4$ and $C_3$ by $R^8$
$\Delta=\begin{vmatrix} a_{r-1} &R^4. a_{r-5} & R^8a_{r-9} \\ a_{r-7} & R^4a_{r-11} &R^8 a_{r-15} \\ a_{r-13} & R^4.a_{r-17} &R^8. a_{r-21} \end{vmatrix}$
But $R^4.a_{r-5}=a_{r-1},R^4.a_{r-11}=a_{r-7},R^4.a_{r-17}=a_{r-13},R^8.a_{r-9}=a_{r-1},R^8.a_{r-15}=a_{r-7},R^8.a_{r-21}=a_{r-13}$
Therefore $\Delta=\begin{vmatrix}a_{r-1} & 1 & 1\\a_{r-7} & 1& 1\\a_{r-13} & 1 & 1\end{vmatrix}$
Since two rows are identical,the value is 0.
Therefore $\Delta=0.$ which is independent of r.
edited Apr 4, 2013