If ${a_1},{a_2},{a_3}, \ldots ,{a_r}$ are in GP, then prove that the determinant
$\left| {\begin{array}{llllllllllllllllllll}{{a_{r + 1}}}&{{a_{r + 5}}}&{{a_{r + 9}}}\\{{a_{r + 7}}}&{{a_{r + 11}}}&{{a_{r + 15}}}\\{{a_{r + 11}}}&{{a_{r + 17}}}&{{a_{r + 21}}}\end{array}} \right|$ is independent of $r$.

We know that, ${a_{r + 1}} = A{R^{(r + 1) - 1}} = A{R^r}$

where $r = r$ th term of a GP, $A =$

First term of a GP and $R =$ Common ratio of GP

We have, $\left| {\begin{array}{llllllllllllllllllll}{{a_{r + 1}}}&{{a_{r + 5}}}&{{a_{r + 9}}}\\{{a_{r + 7}}}&{{a_{r + 11}}}&{{a_{r + 15}}}\\{{a_{r + 11}}}&{{a_{r + 17}}}&{{a_{r + 21}}}\end{array}} \right|$

$= \left| {\begin{array}{cccccccccccccccccccc}{A{R^r}}&{A{R^{r + 4}}}&{A{R^{r + 8}}}\\{A{R^{r + 6}}}&{A{R^{r + 10}}}&{A{R^{r + 14}}}\\{A{R^{r + 10}}}&{A{R^{r + 16}}}&{A{R^{r + 20}}}\end{array}} \right|$
$= A{R^r} \cdot A{R^{r + 6}} \cdot A{R^{r + 10}}\left| {\begin{array}{cccccccccccccccccccc}1&{A{R^4}}&{A{R^8}}\\1&{A{R^4}}&{A{R^8}}\\1&{A{R^6}}&{A{R^{10}}}\end{array}} \right|$

[taking $A{R^r},A{R^{r + 6}}$ and $A{R^{r + 10}}$ common from ${R_1},{R_2}$ and ${R_3}$, respectively]
$= 0$

[since, ${R_1}$ and ${R_2}$ are identicals]