If a_1,a_2,a_3,.........,a_n,...... are in G.P., then the value of the determinant | \beginmatrix \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 \endmatrix |, is

Matrices and Determinants PYQ — JEE Main 2004 · Shift 0 · Q81 with Solution

View written solutionFree

Correct answer: D

Since $a_1,a_2,a_3,\dots$ are in G.P., let $a_k=Ar^{k-1}$ for some constants $A$ and $r$ .
Taking logarithm, $\log a_k=\log A+(k-1)\log r.$ So $\log a_k$ is an arithmetic progression in $k$ .

Let $\alpha=\log A, \qquad \beta=\log r.$ Then $\log a_k=\alpha+(k-1)\beta.$

\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}.$$ Using $\log a_k=\alpha+(k-1)\beta$, the rows become: - First row: $$\big(\alpha+(n-1)\beta,\ \alpha+n\beta,\ \alpha+(n+1)\beta\big)$$ - Second row: $$\big(\alpha+(n+2)\beta,\ \alpha+(n+3)\beta,\ \alpha+(n+4)\beta\big)$$ - Third row: $$\big(\alpha+(n+5)\beta,\ \alpha+(n+6)\beta,\ \alpha+(n+7)\beta\big)$$ 4. Observe that each row is obtained from the previous row by adding the same vector $(3\beta,3\beta,3\beta)$: $$R_2-R_1=(3\beta,3\beta,3\beta),$$ $$R_3-R_2=(3\beta,3\beta,3\beta).$$ Hence, $$R_3-2R_2+R_1=(0,0,0).$$ So the three rows are linearly dependent. Therefore the determinant must be zero: $$D=0.$$ 5. Checking options: - A: $-2$ ❌ - B: $1$ ❌ - C: $2$ ❌ - D: $0$ ✅ Hence the correct answer is $\boxed{0}$.

View written solutionFree

Correct answer: D

Since $a_1,a_2,a_3,\dots$ are in G.P., let $a_k=Ar^{k-1}$ for some constants $A$ and $r$ .
Taking logarithm, $\log a_k=\log A+(k-1)\log r.$ So $\log a_k$ is an arithmetic progression in $k$ .

Let $\alpha=\log A, \qquad \beta=\log r.$ Then $\log a_k=\alpha+(k-1)\beta.$

\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}.$$ Using $\log a_k=\alpha+(k-1)\beta$, the rows become: - First row: $$\big(\alpha+(n-1)\beta,\ \alpha+n\beta,\ \alpha+(n+1)\beta\big)$$ - Second row: $$\big(\alpha+(n+2)\beta,\ \alpha+(n+3)\beta,\ \alpha+(n+4)\beta\big)$$ - Third row: $$\big(\alpha+(n+5)\beta,\ \alpha+(n+6)\beta,\ \alpha+(n+7)\beta\big)$$ 4. Observe that each row is obtained from the previous row by adding the same vector $(3\beta,3\beta,3\beta)$: $$R_2-R_1=(3\beta,3\beta,3\beta),$$ $$R_3-R_2=(3\beta,3\beta,3\beta).$$ Hence, $$R_3-2R_2+R_1=(0,0,0).$$ So the three rows are linearly dependent. Therefore the determinant must be zero: $$D=0.$$ 5. Checking options: - A: $-2$ ❌ - B: $1$ ❌ - C: $2$ ❌ - D: $0$ ✅ Hence the correct answer is $\boxed{0}$.