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Question: Answered & Verified by Expert
If $a_{1} a_{2} a_{3} \ldots a_{9}$ are in $\mathrm{AP}$, then the value of $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is
MathematicsLimitsKCETKCET 2020
Options:
  • A $\frac{9}{2}\left(a_{1}+a_{9}\right)$
  • B $\left(a_{1}+a_{9}\right)$
  • C $\log _{e}\left(\log _{e} e\right)$
  • D 1
Solution:
2776 Upvotes Verified Answer
The correct answer is: $\log _{e}\left(\log _{e} e\right)$
$a_{1}, a_{2}, a_{3}, \ldots, a_{9}$ are in AP.
$\left|\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}\right|$
Apply $R_{1} \rightarrow R_{1}+R_{2}-2 R_{2}$
$\left|\begin{array}{ccc}
a_{1}+a_{7}-2 a_{4} & a_{2}+a_{8}-2 a_{5} & a_{3}+a_{9}-2 a_{6} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}\right|$
$\left|\begin{array}{ccc}
0 & 0 & 0 \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}\right|$
$\begin{aligned}
&{\left[\because a_{1}+a_{7}-2 a_{4}=a_{1}+a_{1}+6 d-2 a_{1}-6 d=0\right.} \\
&\text { Similarly } \left.a_{2}+a_{4}-2 a_{5}=0, a_{3}+a_{9}-2 a_{6}=0\right] \\
&=0 \\
&=\log _{e} 1=\log _{e}\left(\log _{e} e\right)
\end{aligned}$

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