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यदि ${a_1},{a_2},{a_3}.....{a_n}....$ गुणोत्तर श्रेणी में हैं, तब सारणिक $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$ का मान होगा
$-2$
$1$
$2$
$0$
Solution
(d) दिया है, ${a_1},\,{a_2},\,{a_3}$….. $an$ गुणोत्तर श्रेणी में हैं
तब $r = \frac{{{a_2}}}{{{a_1}}}$ अर्थात् $r = \frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = …..$..
अत: $\log r = \log ({a_{n + 1}}) – \log ({a_n}) = \log ({a_{n + 2}}) – \log ({a_{n + 1}}) = …..$
अब $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$
संक्रिया ${C_2} \to {C_2} – {C_1}$ और ${C_3} \to {C_3} – {C_2}$
= $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{(\log {a_{n + 1}} – \log {a_n})}&{(\log {a_{n + 2}} – \log {a_{n + 1}})}\\{\log {a_{n + 3}}}&{(\log {a_{n + 4}} – \log {a_{n + 3}})}&{(\log {a_{n + 5}} – \log {a_{n + 4}})}\\{\log {a_{n + 6}}}&{(\log {a_{n + 7}} – \log {a_{n + 6}})}&{(\log {a_{n + 8}} – \log {a_{n + 7}})}\end{array}\,} \right|$
= $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log r}&{\log r}\\{\log {a_{n + 3}}}&{\log r}&{\log r}\\{\log {a_{n + 6}}}&{\log r}&{\log r}\end{array}\,} \right|{\rm{ }} = {\rm{ 0}}$.
