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  • I have a doubt, kindly clarify. - Sequence and series - JEE Main

If\; \; a_{1},a_{2},a_{3},............,a_{n}...are\; in\; G.P., then\; the \; determinant

 

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best_answer

As we learnt in 

General term of a GP -

T_{n}= ar^{n-1}
 

- wherein

a\rightarrow first term

r\rightarrow common ratio

 

Given a1, a2,  a3,  ...an  are in A.P

 

\Delta = \begin{vmatrix} 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{vmatrix}

\therefore a n  =a1 rn-1 ,  a n+2  =a1 rn+1

a n+1  =a1 rn ,  a n+3  =a1 rn+2

\therefore  a n+8=a1 rn+7

\Delta =\begin{vmatrix} log\:a_{1}r^{n-1}\: log\:a_{1}r^{n}\:log\:a_{1}r^{n+1} & & \\ log\:a_{1}r^{n+2}\: log\:a_{1}r^{n+3}\:log\:a_{1}r^{n+4} & & \\ log\:a_{1}r^{n+5}\: log\:a_{1}r^{n+6}\:log\:a_{1}r^{n+7} & & \end{vmatrix}

\Delta =\begin{vmatrix} log\:a_{1}+({n-1})log r\ \;log\: a_{1}+nlogr\ \; log\:a_{1}(n+1)log\:r & & \\ log\:a_{1}+(n+2)logr\ \: log\:a_{1}+(n+3)\:logr \ log a_{1}+(n+4)logr & & \\ log\:a_{1}+(n+5)logr\ \: log\:a_{1}(n+6)logr\ \:loga_{1}(n+7)logr & & \end{vmatrix}

\Delta= c2 + c2 -c1 & c3 + c3-c2

\begin{vmatrix} log\ a_{1}+(n-1)log r \ log r \ \ logr & & \\ log\ a_{1}+(n+2)log r \ log r \ \ logr & & \\ log\ a_{1}+(n+3)log r \ log r \ \ logr & & \end{vmatrix}=0

Since two columns are equal


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