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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)

\\\vspace{\baselineskip} a\textsubscript{1}, a\textsubscript{2}$ \ldots $ , a\textsubscript{r} are in G.P\\ \vspace{\baselineskip} \text{We know that, } a\textsubscript{r+1} = AR\textsuperscript{(r+1)-1} = AR\textsuperscript{r} $ \ldots $ (i)\\ \\ \vspace{\baselineskip}[$\because$ a\textsubscript{n} = ar\textsuperscript{n-1}, \text{where a = first term and r = common ratio}]\\ \vspace{\baselineskip}

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[f r o m(i)]}$

\\Taking AR", \\\mathrm{AR}^{\mathrm{r}+6}$ and $\mathrm{AR}^{\mathrm{r}+10}$ common from $\mathrm{R}_{1}, \mathrm{R}_{2}$ and $\mathrm{R}_{3}$

\\\\ =\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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