Question

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

Answer 1

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, $\left|\begin{array}{ccc}{a}_{r+1}& a_{r+5}& a_{r+9}\\ a_{r+7}& a_{r+11}& a_{r+15}\\ a_{r+11}& a_{r+17}& a_{r+21}\end{array}\right|=\left|\begin{array}{ccc}A{R}^r& A{R}^{r+4}& A{R}^{r+8}\\ A{R}^{r+6}& A{R}^{r+10}& A{R}^{r+14}\\ A{R}^{r+10}& A{R}^{r+16}& A{R}^{r+20}\end{array}\right|=A{R}^r.A{R}^{r+6}.A{R}^{r+10}\left|\begin{array}{ccc}1& A{R}^4& A{R}^8\\ 1& A{R}^4& A{R}^8\\ 1& A{R}^6& A{R}^{10}\end{array}\right|$
[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]