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If $a_{n}(>0)$ be the $n^{\text {th }}$ term of a G.P. then $\left|\begin{array}{lll}\log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{array}\right|$ is equal to
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The correct answer is:
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$\log a_{n}=\log \left(a r^{n-1}\right)=\log a+(n-1) \log r$
$=\left|\begin{array}{ccc}\log \mathrm{a}+(\mathrm{n}-1) \log \mathrm{r} & \log \mathrm{a}+\mathrm{nlogr} & \log \mathrm{a}+(\mathrm{n}+1) \log \mathrm{r} \\ 3 \log \mathrm{r} & 3 \log \mathrm{r} & 3 \log \mathrm{r} \\ 6 \log r & 6 \log r & 6 \log r\end{array}\right|$
= 0
$=\left|\begin{array}{ccc}\log \mathrm{a}+(\mathrm{n}-1) \log \mathrm{r} & \log \mathrm{a}+\mathrm{nlogr} & \log \mathrm{a}+(\mathrm{n}+1) \log \mathrm{r} \\ 3 \log \mathrm{r} & 3 \log \mathrm{r} & 3 \log \mathrm{r} \\ 6 \log r & 6 \log r & 6 \log r\end{array}\right|$
= 0
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