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  • If a1, a2, a3, ..., ar are in G.P., then prove that the determinant is independent of r.

If a1, a2, a3, ..., ar are in G.P., then prove that the determinant \left|\begin{array}{ccc} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{array}\right|  is independent of r.

Answers (1)

 

\begin{align*} a_1, a_2, \ldots , a_r \text{ are in G.P} \\ \text{We know that, } a_{r+1} = AR^{(r+1)-1} = AR^r \ldots (i) \\ \\ \text{[$\because$ } a_n = ar^{n-1}, \text{where } a \text{ = first term and } r \text{ = common ratio]} \\ \end{align*}

A is the first term of G.P

R is the common ratio of G.P.

 

\therefore \left|\begin{array}{lll}a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21}\end{array}\right| = \left|\begin{array}{ccc}A R^{r} & A R^{r+4} & A R^{r+8} \\ A R^{r+6} & A R^{r+10} & A R^{r+14} \\ A R^{r+10} & A R^{r+16} & A R^{r+20}\end{array}\right| \ldots \text{[from (i)]}

 

\text{Taking } AR^r, AR^{r+6} \text{ and } AR^{r+10} \text{ common from } R_1, R_2 \text{ and } R_3

\\\\ =\mathrm{AR}^{\mathrm{r}} \times \mathrm{AR}^{\mathrm{r}+6} \times \mathrm{AR}^{\mathrm{r}+10}\left|\begin{array}{ccc} 1 & \mathrm{AR}^{4} & \mathrm{AR}^{8} \\ 1 & \mathrm{AR}^{4} & \mathrm{AR}^{8} \\ 1 & \mathrm{AR}^{6} & \mathrm{AR}^{10} \end{array}\right|

Whenever any the values in any two rows or columns of a determinant are identical, the resultant value of that determinant is 0
Rows 1 and 2 are identical

\\ \therefore\left|\begin{array}{lll}a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21}\end{array}\right|=0\\

Hence Proved

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