# If a1, a2, a3, …, ar are in G.P.,

Question:

If a1a2a3, …, ar are in G.P., then prove that the determinant

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

is independent of r.

Solution:

We know that,

$a_{r+1}=A R^{(r+1)-1}=A R^{r}$

where $a_{r}=r$ th term of G.P.,

$A=$ First term of G.P.

and $R=$ Common ratio of G.P.

Now,

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

[Taking $A R^{r}, A R^{r+6}$ and $A R^{r-10}$ common from $R_{1}, R_{2}$ and $R_{3}$, respectively]

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

$=0$ [As $R_{1}$ and $R_{2}$ are identical $]$

Hence, the determinant is independent of r.