JEE MAIN - Mathematics (2004 - No. 15)
$$\left| {\matrix{
{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr
{\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr
{\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr
} } \right|,$$ is
$$-2$$
$$1$$
$$2$$
$$0$$
Explanation
$$\left| {\matrix{
{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr
{\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr
{\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr
} } \right|$$
$$ = \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr } } \right|$$
$$ = \left| {\matrix{ {\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr } } \right|$$
$$ = 0\left[ \, \right.$$ Apply $$\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$$
$$ = \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr } } \right|$$
$$ = \left| {\matrix{ {\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr } } \right|$$
$$ = 0\left[ \, \right.$$ Apply $$\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$$
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