If 2^{a_1}, 2^{a_2}, 2^{a_3}, dots, 2^{a_n} a | vedclass

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(C) Given that $2^{a_1}, 2^{a_2}, 2^{a_3}, \dots$ are in $G.P.$This implies that the exponents $a_1, a_2, a_3, \dots$ are in $A.P.$Let the common diff

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(C) Given that $2^{a_1}, 2^{a_2}, 2^{a_3}, \dots$ are in $G.P.$This implies that the exponents $a_1, a_2, a_3, \dots$ are in $A.P.$Let the common difference of the $A.P.$ be $d$. Then $a_k = a_1 + (k-1)d$.In the determinant $\Delta = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \ a_{n+1} & a_{n+2} & a_{n+3} \ a_{2n+1} & a_{2n+2} & a_{2n+3} \end{array} ight|$,notice that the columns are in $A.P.$Specifically,$a_2 - a_1 = d$,$a_{n+2} - a_{n+1} = d$,and $a_{2n+2} - a_{2n+1} = d$.Also,$a_3 - a_2 = d$,$a_{n+3} - a_{n+2} = d$,and $a_{2n+3} - a_{2n+2} = d$.Applying the column operation $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_2$,we get columns of constant differences.Since the rows are also in $A.P.$,the rows $R_1, R_2, R_3$ satisfy $R_1 + R_3 = 2R_2$.Thus,$R_2 = \frac{R_1 + R_3}{2}$,which means the rows are linearly dependent.Therefore,the determinant $\Delta = 0$.

If $2^{a_1}, 2^{a_2}, 2^{a_3}, \dots, 2^{a_n}$ are in $G.P.$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_{n+1} & a_{n+2} & a_{n+3} \\ a_{2n+1} & a_{2n+2} & a_{2n+3} \end{array} \right|$ is equal to

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