Question
If $a_1, a_2, a_3, ...,$ ar are in G.P., then prove that the determinant $\begin{bmatrix}\text{a}_{\text{r}+1}&\text{a}_{\text{r}+5}&\text{a}_{\text{r}+9}\\\text{a}_{\text{r}+7}&\text{a}_{\text{r}+11}&\text{a}_{\text{r}+15}\\\text{a}_{\text{r}+11}&\text{a}_{\text{r}+17}&\text{a} _{\text{r}+21}\end{bmatrix}$ is independent of r.

Answer

We know that, $\text{a}_{\text{r}+1}=\text{AR}^{(\text{r}+1)}=\text{AR}^\text{r}$ where $r = r^{th}$ term of a GA, A = First term of a GP and R = Common ratio of GP We have $\begin{vmatrix}\text{a}_{\text{r}+1}&\text{a}_{\text{r}+5}&\text{a}_{\text{r}+9}\\\text{a}_{\text{r}+7}&\text{a}_{\text{r}+11}&\text{a}_{\text{r}+15}\\\text{a}_{\text{r}+11}&\text{a}_{\text{r}+17}&\text{a} _{\text{r}+21}\end{vmatrix}$ $=\text{AR}^{\text{r}}\text{AR}^{\text{r}+6}\text{AR}^{\text{r}+10}\begin{vmatrix}1&\text{AR}^4&\text{AR}^8\\1&\text{AR}^4&\text{AR}^8\\1&\text{AR}^6&\text{AR}^{10}\end{vmatrix}$[Taking $\text{AR}^\text{r}.\text{AR}^{\text{r}+6}.\text{AR}^{\text{r}+10}$ common from $\text{R}_1,\text{R}_2\text{ and R}_2$ respectively]
$=0\ [\text{Since, R}_1\text{ and R}_2\text{ are identicals}]$

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