If a_{n} (0) is the n^{text{th}} term of a G | vedclass

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(D) Let the $G$.$P$. be $a, ar, ar^2, \dots$. Then the $n^{	ext{th}}$ term is $a_n = ar^{n-1}$.Taking logarithm, we have $\log a_n = \log a + (n-1) \

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$0$

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(D) Let the $G$.$P$. be $a, ar, ar^2, \dots$. Then the $n^{	ext{th}}$ term is $a_n = ar^{n-1}$.Taking logarithm, we have $\log a_n = \log a + (n-1) \log r$.Let $A = \log a$ and $D = \log r$. Then $\log a_n = A + (n-1)D$.The determinant becomes:$\Delta = \left|\begin{array}{lll} A+(n-1)D & A+nD & A+(n+1)D \ A+(n+2)D & A+(n+3)D & A+(n+4)D \ A+(n+5)D & A+(n+6)D & A+(n+7)D \end{array}ight|$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$:$R_2 - R_1 = \begin{bmatrix} 3D & 3D & 3D \end{bmatrix}$$R_3 - R_2 = \begin{bmatrix} 3D & 3D & 3D \end{bmatrix}$Since two rows ($R_2$ and $R_3$) are identical, the value of the determinant is $0$.

If $a_{n} (>0)$ is the $n^{\text{th}}$ term of a $G$.$P$.,then the value of the determinant $\left|\begin{array}{lll}\log a_{n} & \log a_{n+1} & \log a_{n+2} \\ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \\ \log a_{n+6} & \log a_{n+7} & \log a_{n+8}\end{array}\right|$ is equal to:

Similar Questions

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